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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`intx"cosec"^(2)xdx=?`A. `xcotx-log|sinx|+C`B. `-xcotx+log|sinx|+C`C. `xtanx-log|secx|+C`D. none of these |
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Answer» Correct Answer - B `intunderset(I)(x)underset(II)("cosec"^(2))xdx` |
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| 2. |
Evaluate the following integrals: `int"cosec"^(3)xdx` |
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Answer» Correct Answer - `-(1)/(2)"cosec "xcotx+(1)/(2)log|"tan"(x)/(2)|+C` `I=int"cosec"("cosec"^(2)x)dx`. Now, integrate by parts. |
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| 3. |
`int((1+sinx))/((1+cosx))dx=?`A. `"tan"(x)/(2)+2log|"cos"(x)/(2)|+C`B. `"-tan"(x)/(2)+2log|"cos"(x)/(2)|+C`C. `"tan"(x)/(2)-2log|"cos"(x)/(2)|+C`D. none of these |
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Answer» Correct Answer - C `I=int(1)/((1+cosx))dx+int(sinx)/((1+cosx))dx`. `int(dx)/(2cos^(2)((x)/(2)))+int(2sin((x)/(2))cos((x)/(2)))/(2cos^(2)((x)/(2)))dx` `=(1)/(2)sec^(2)(x)/(2)dx+int"tan"(x)/(2)dx` `=intsec^(2)tdt+2tantdt," where "(x)/(2)=t` `=tant-2log|cost|+C="tan"(x)/(2)-2log|"cos"(x)/(2)|+C`. |
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| 4. |
`int((x^(2)+1))/((x^(4)+1))dx=?`A. `(1)/(sqrt(2))tan^(-1)(x-(1)/(x))+C`B. `(1)/(sqrt(2))cot^(-1){(x-(1)/(x))}+C`C. `(1)/(sqrt(2))tan^(-1){(1)/(sqrt(2))(x-(1)/(x))}+C`D. none of these |
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Answer» Correct Answer - C On dividing the numerator and denominator by `x^(2)`, we get `I=int((1+(1)/(x^(2))))/((x^(2)+(1)/(x^(2))))dx=int((1+(1)/(x^(2))))/({(x-(1)/(x))^(2)+2})dx` `=int(dt)/((t^(2)+2))," where "(x-(1)/(x))=t and(1+(1)/(x^(2)))dx=dt` `int(dt)/([t^(2)+(sqrt(2))^(2)])=(1)/(sqrt(2))+C=(1)/(sqrt(2))tan^(-1){(1)/(sqrt(2))(x-(1)/(x))}+C`. |
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| 5. |
`int(sqrt((2+logx)))/(x)dx` |
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Answer» Correct Answer - `(2)/(3)(2+logx)^(3//2)+C` Put (2x+logx)=t. |
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| 6. |
`int"cosec"xdx=?`A. `log|"cosec"x-cotx|+C`B. `-log|"cosec"x-cotx|+C`C. `log|"cosec "x+cotx|+C`D. none of these |
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Answer» Correct Answer - A `I=int("cosec "x("cosec "x-cotx))/(("cosec "x=cotx))dx=int(1)/(t)dt," where " ("cosec" x-cot x)=t`. `=log|t|+C=log|"cosec "x-cotx|+C`. |
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| 7. |
`int((1+tanx)/(1-tanx))dx` |
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Answer» Correct Answer - `-log|cosx-sinx|+C` `I=int((cosx+sinx))/((cosx-sinx))dx.Put(cosx-sinx)=t`. |
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| 8. |
`int(sec^(2)x)/(sqrt(1-tan^(2)x))dx=?`A. `sin^(-1)(tanx)+C`B. `cos^(-1)(sinx)+C`C. `tan^(-1)(cosx)+C`D. `tan^(-1)(sinx)+C` |
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Answer» Correct Answer - A Put `tanx=t andsec^(2)xdx=dt`. `:." "I=int(dt)/(sqrt(1-t^(2)))=sin^(-1)(t)+C=sin^(-1)(tanx)+C`. |
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| 9. |
`int(e^x(1+x))/(cos^2(x e^x)) dx=``2(log)_ecos(x e^x)+C`(b) `sec(x e^x)+C`(c) `tan(x e^x)+C`(d) `tan(x+e^x)+C`A. `tan(xe^(x))+C`B. `cot(xe^(x))+C`C. `ex^(x)tanx+C`D. none of these |
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Answer» Correct Answer - A Put `xe^(x)=t and(xe^(x)+e^(x))dx=dt`. `:." "I=int(1)/(cos^(2)t)dt=intsec^(2)"t dt"=tan t+C`. |
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| 10. |
`intsecxdx=?`A. `log|secx-tanx|+C`B. `-log|secx+tanx|+C`C. `log|secx+tanx|+C`D. none of these |
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Answer» Correct Answer - C `I=int(secx(secx+tanx))/((secx+tanx))dx=int(1)/(t)`dt, where (secx+tanx)=t. `=log|t|+C=log|secx+tanx|+C`. |
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| 11. |
(i) `int((1+tanx))/((x+logsecx))dx` , (ii) `int((1-sin2x))/((x+cos^(2)x))` |
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Answer» Correct Answer - (i) `log|x+log(secx)|+C` , (ii) `log|x+cos^(2)x|+C` to 53. In each these questions, put the denominator equal to t. |
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| 12. |
`inttanxdx=?`A. `log|cosx|+C`B. `-log|cosx|+C`C. `log|sinx|+C`D. `-log|sinx|+C` |
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Answer» Correct Answer - B `I=int(sinx)/(cosx)dx=int(-dt)/(t)`, where cos x=t. `=-log|t|+C=-log|cosx|+C`. |
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| 13. |
`int(sec^(2)x)/(sqrt(1-tan^(2)x))dx=?` |
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Answer» Correct Answer - `sin^(-1)(tanx)+C` Put tanx=t. |
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| 14. |
`inte^(-x)" cosec"^(2)(2e^(-x)+5)dx` |
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Answer» Correct Answer - `(1)/(2)cot(2e^(-x)+5)+C` Put `(2e^(-x)+5)=t`. |
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| 15. |
`intsinxcosdx` |
| Answer» Correct Answer - `(1)/(2)sin^(2)x+C` | |
| 16. |
`int(dx)/(sqrt(2x+3)+sqrt(2x-3))`A. `(1)/(18)(2x+3)^((3)/(2))+(1)/(18)(2x-3)^((3)/(2))+C`B. `(1)/(18)(2x+3)^((3)/(2))-(1)/(18)(2x-3)^((3)/(2))+C`C. `(1)/(12)(2x+3)^((3)/(2))-(1)/(12)(2x-3)^((3)/(2))+C`D. none of these |
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Answer» Correct Answer - B `I=int(1)/({sqrt(2x+3)+sqrt(2x-3)})xx({sqrt(2x+3)-sqrt(2x-3)})/({(2x+3)-sqrt(2x-3)})dx`. `=int({sqrt(2x+3)-sqrt(2x-3)})/({(2x+3)-(2x-3)})dx=(1)/(6)int{sqrt(2x+3)-sqrt(2x-3)}dx` `=(1)/(18)(2x+3)^((3)/(2))-(1)/(18)(2x+3)^((3)/(2))+C`. |
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| 17. |
`int x^5/sqrt(1+x^3) dx` |
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Answer» Correct Answer - `(2)/(9)(1+x^(3))^(3//2)-(2)/(3)(1+x^(3))^(1//2)+C` `Put(1+x^(3))=t^(2)." Then",I=(2)/(3)int(t(t^(2)-1))/(sqrt(t))dt=(2)/(3)int(t^(5//2)-t^(1//2))dt`. |
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| 18. |
`int(dx)/((1-tanx))=?` |
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Answer» Correct Answer - `(1)/(2)x-(1)/(2)log|sinx-cosx|+C` `I=int(-cosx)/(sinx-cosx)dx=int((sinx-cosx)-(sinx+cosx))/(2(sinx-cosx))dx`. |
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| 19. |
`int(cos2x)/((sinx+cosx)^(2))dx` is equal to |
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Answer» Correct Answer - `log|sinx+cosx|+C` `I=int((cos^(2)x-sin^(2)x))/((sinx+cosx)^(2))dx=int((cosx-sinx))/((sinx+cosx))dx`. |
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| 20. |
Evaluate:`int2xsec^3(x^2+3)tan(x^2+3)dx` |
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Answer» Correct Answer - `(sec^(3)(x^(2)+3))/(3)+c` Put `sec(x^(2)+3)=t`. |
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| 21. |
`intx^(2)e^(x^(3))cos(e^(x^(3)))dx` is equalto |
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Answer» Correct Answer - `(1)/(3)sin(e^(x^(3)))+C` Put `e^(x^(3))=t`. |
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| 22. |
`int((x+1)(x+logx)^(2))/(x)dx=?` |
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Answer» Correct Answer - `(1)/(3)(x+logx^(3))+C` `I=int(1+(1)/(x))(x+logx)^(2)dx." "Put(x+logx)=t`. |
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| 23. |
`int(sinx)/((sinx-cosx))dx=?` |
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Answer» Correct Answer - `(1)/(2)x+(1)/(2)log|sinx-cosx|+C` `I=int((sinx-cosx)+(sinx+cosx))/(2(sinx-cosx))dx`. |
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| 24. |
`int(3x^2)/(1+x^6)dx`A. `sin^(-1)x^(3)+C`B. `cos^(-1)x^(3)+C`C. `tan^(-1)x^(3)+C`D. `cot^(-1)x^(3)+C` |
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Answer» Correct Answer - C Putting `x^(3)=t and3x^(2)dx=dt`, we get : `I=int(dt)/((1+t^(2)))=tan^(-1)t+C=tan^(-1)x^(3)+C`. |
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| 25. |
`int (2x+1)(sqrt(x^(2)+x+1))dx`A. `(3)/(2)(x^(2)+x+1)^((3)/(2))+C`B. `(2)/(3)(x^(2)+x+1)^((3)/(2))+C`C. `(3)/(2)(2x+1)^(3//2)+C`D. none of these |
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Answer» Correct Answer - B Put `x^(2)+x+1=t and(2x+1)dx=dt`. , `I=intsqrt(t)dt=(t^((3)/(2)))/(((3)/(2)))+C=(2)/(3)(x^(2)+x+1)^((3)/(2))+C`. |
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| 26. |
`int"cosec"^(2)(2x+5)dx` |
| Answer» Correct Answer - `-(1)/(2)cot(2x+5)+C` | |
| 27. |
`int(x)/(sqrt(1+x))dx` |
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Answer» Correct Answer - `(2)/(3)(1+x)^(3//2)-2sqrt(1+x)+C` `Put(1+x)=t^(2)`. |
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| 28. |
`int(sin2x)/(1+bcosx)^(2)dx` |
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Answer» Correct Answer - `(-2)/(b^(2)){:[log|a+bcosx|+(a)/((a+bcosx))]:}+C` Put `(a+bcosx)=t."Then",cosx=((t-a))/(b)and-sin"x dx"=(1)/(b)dt`. `:.I=-(2)/(b)*int(1)/(t^(2))((t-a)/(b))dt=-(2)/(b^(2))*int{(1)/(t)-(a)/(t^(2))}dt`. |
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| 29. |
`int((cosx-sinx))/((1+sin2x))dx` |
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Answer» Correct Answer - `(-1)/((sinx+cosx))+C` `I=int((cosx-sinx))/((sinx+cosx)^(2))dx.Put(sinx+cosx)=t`. |
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| 30. |
int(dx)/(xsqrt(x^(6)-1))=?`A. `(1)/(3)sec^(-1)x^(3)+C`B. `(1)/(3)"cosec"^(-1)x^(3)+C`C. `(1)/(3)cot^(-1)x^(3)+C`D. none of these |
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Answer» Correct Answer - A `I=int(x^(2))/(x^(3)sqrt(x^(6)-1))dx`. `=(1)/(3)int(dt)/(tsqrt(t^(2)-2))`, where `x^(3)=t` `=(1)/(3)sec^(-1)t+C=(1)/(3)sec^(-1)x^(3)+C`. |
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| 31. |
`int 1/(xsqrt(x^4-1))dx` |
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Answer» Correct Answer - `(1)/(2)sec^(-1)x^(2)+C` Multiply num. and denom. By x and Put `x^(2)=t`. |
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| 32. |
`int((logx)^(2))/(x)dx.` |
| Answer» Correct Answer - `(1)/(3)(logx)^(3)+C` | |
| 33. |
`int("cosec"^(2)(logx))/(x)dx` |
| Answer» Correct Answer - `-cot(logx)+C` | |
| 34. |
`int(1)/(sin^(2)xcos^(2)x)dx=?`A. `tanx+cotx+C`B. `tanx-cotx+C`C. `-tanx+cotx+C`D. none of these |
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Answer» Correct Answer - B `I=it((sin^(2)x+cos^(2)x))/(sin^(2)xcos^(2)x)dx=int(sin^(2)x)/(sin^(2)xcos^(2)x)dx+int(cos^(2)x)/(sin^(2)xcos^(2)x)dx`. `=int(1)/(cos^(2)x)dx+int(1)/(sin^(2)x)dx=intsec^(2)xdx+int"cosec"^(2)xdx=tanx-cotx+C`. |
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| 35. |
Evaluate : (i) `int(3x^(2))/((1+x^(6)))dx` (ii) `int(x^(3))/((x^(2)+1)^(3))dx` (iii) `(x^(8))/((1-x^(3))^(1//3))dx` |
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Answer» (i) Put `x^(3)=t` so that `3x^(2)dx=dt`. `:.int(3x^(2))/((1+x^(6)))dx=int(dt)/((1+t^(2)))=tan^(-1)t+C=tan^(-1)x^(3)+C`. (ii) Put `(x^(2)+1)` =t so that `x^(2)=(t-1)andxdx=(1)/(2)dt`. `:.int(x^(3))/((x^(2)+1)^(3))dx=int(x^(2)*x)/((x^(2)+1))dx` `=(1)/(2)int((t-1))/(t^(3))dt=(1)/(2)int(1)/(t^(2))dt-(1)/(t^(3))dt` `=(-1)/(2t)+(1)/(4t^(2))+C=(-1)/(2(x^(2)+1))+(1)/(4(x^(2)+1)^(2))+C` `=(-1+2x^(2))/(4(x^(2)+1)^(2))+C`. (iii) Put `(1=x^(3))=t` so that `x^(3)=(1-t)andx^(2)dx=-(1)/(3)dt`. `:.int(x^(8))/((1-x^(3))^(1//3))dx=int(x^(6)*x^(2))/((1-x^(3))^(1//3))dx` `=-(1)/(3)int((1-t)^(2))/(t^(1//3))dt=-(1)/(3)((1+t^(2)-2t))/(t^(t//3))dt` `=-(1)/(3)intt^(-1//3)dt-(1)/(3)intt^(5//3)dt+(2)/(3)intt^(2//3)dt` `=-(1)/(2)t^(2//3)-(1)/(8)t^(8//3)+(2)/(5)t^(5//3)+C` `=-(1)/(2)(1-x^(3))^(2//3)-(1)/(8)(1-x^(3))^(8//3)+(2)/(5)(1-x^(3))^(5//3)+C`. |
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| 36. |
`intx^(2)sinx^(3)dx=?` |
| Answer» Correct Answer - `-(1)/(3)cosx^(3)+C` | |
| 37. |
`int(1)/(xcos^(2)(1+logx))dx=?` |
| Answer» Correct Answer - `tan(1+logx)+C` | |
| 38. |
Evaluate `int (dx)/(xsqrt(x^6-1))` |
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Answer» Put `x^(3)=t` so that `3x^(2)dx=dt" or "x^(2)dx=(1)/(3)dt`. `:.int(dx)/(x*sqrt(x^(6)-1))dx` [multiplying numerator and denominator by `x^(2)]` `=(1)/(3)int(1)/(tsqrt(t^(2)-1))dt=(1)/(3)sec^(-1)t+C=(1)/(3)sec^(-1)t+C=(1)/(3)sec^(-1)x+C`. |
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| 39. |
Evaluate:`int1/(sqrt(x)+x)dx`A. `log|1+sqrt(x)|+C`B. `2log|1+sqrt(x)|+C`C. `(1)/(sqrtx)tan^(-1)sqrt(x)+C`D. none of these |
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Answer» Correct Answer - B `I=int(dx)/(sqrt(x)(1+sqrt(x)))*"Put"(1+sqrtx)=t and(1)/(2sqrt(x))dx=dt`. `:." "I=2*int(1)/(t)dt=2log|t|+C`. |
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| 40. |
Evaluate:`int1/(sqrt(x)+x)dx` |
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Answer» `int(1)/((sqrt(x)+x))dx=int(1)/(sqrt(x)(1+sqrt(x)))dx` Now, put `(1+sqrt(x))` =t so that `(1)/(sqrt(x))dx=2dt`. `:.int(1)/(sqrt(x)+x)dx=int(1)/(sqrt(x)(1+sqrt(x)))dx` `=2intdt=2log|t|+C=2log|{:(1+sqrt(x)):}|+C`. |
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| 41. |
`intxsqrt(3x-2)dx` |
| Answer» Correct Answer - `(2)/(45)(3x-2)^(5//2)+(4)/(27)(3x-2)^(3//2)+C` | |
| 42. |
`intxsqrt(x^(2)-x)dx=?`A. `(1)/(3)(x^(2)-1)^((3)/(2))+C`B. `(2)/(3)(x^(2)-1)^((3)/(2))+C`C. `(1)/(sqrt(x^(2)-1))+C`D. none of these |
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Answer» Correct Answer - A Put `x^(2)-1=t and2xdx=dt`. `I=(1)/(2)sqrt(t)dt=(1)/(2)*(t^((3)/(2)))/((3)/(2))+C=(1)/(3)(x^(2)-1)^((3)/(2))+C`. |
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| 43. |
`intxsqrt(x-1)dx=?`A. `(2)/(3)(x-1)^((3)/(2))+C`B. `(2)/(5)(x-1)^((5)/(2))+C`C. `(2)/(5)(x-1)^((5)/(2))+(3)/(2)(x-1)^((3)/(2))+C`D. none of these |
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Answer» Correct Answer - C `I=int{(x-1+1)sqrt(x-1)}dx=int{(x-1)^((3)/(2))+(x-1)^((1)/(2))}dx`. `=(2)/(5)(x-1)^((5)/(2))+(3)/(2)(x-1)^((3)/(2))+C`. |
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| 44. |
Evaluate the following integrals: `int5^(5^(5^(x)))*5^(5^(x))*5^(x)dx` |
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Answer» Correct Answer - `(5^(5^(5^(x))))/((log5)^(3))+C` Putting `5^(x)=thArr5^(x)log5dx=dthArr5^(x)dx=(1)/(log5)dt`. `:.I=int5^(5^t)5^(t)(1)/(log5)dt=int5^(u)*(1)/((log5)^(2))"du, where "5^(t)=uhArr5^(t)log5dt=du` `=int(1)/((log5)^(3))"du, where "5^(u)=vhArr5^(u)log5du=dv` `(v)/((log5)^(3))+C(5^(u))/((log5)^(3))+C=(5^(5^(t)))/((log5)^(3))+C=(5^(5^(5^(x))))/((log5)^(3))+C`. |
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| 45. |
Evaluate : (i) `int((x-1))/(sqrt(x+4))dx` (ii) `intxsqrt(x+2)dx` (iii) `f(4x-2)sqrt(x^(2)+x++1)dx` (iv) `int((4x+3))/(sqrt(2x^(2)+3x+1))dx` |
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Answer» (i) Put `(x+4)=t^(2)` so that `x=(t^(2)-4)anddx=2tdt`. `:.int((x-1))/(sqrt(x+4))dx=2int((t^(2)-5))/(t)dt` `=2intt^(2)dt-10intdt=(2t^(3))/(3)-10t+C` `=(2)/(3)(x+4)^(3//2)-10(x+4)^(1//2)+C`. (ii) Put `(x+2)=t^(2)` so that `x=(t^(2)-2)anddx=2tdt`. `:.intxsqrt(x+2)dx=int(t^(2)-2)2dt=2intt^(4)dt-4f^(2)dt` `=(2t^(5))/(5)-(4t^(3))/(3)+C=(2(x+2)^(5//2))/(5)-(4(x+2)^(3//2))/(3)+C`. (iii) Put `(x^(2)+x+1)=t` so that (2x+1)dx=dt. `:.int(4x+2)(sqrt(x^(2)+x+1))dx=2intsqrt(t)dt` `=(4)/(3)t^(3//2)+C=(4)/(3)(x^(2)+x+1)^(3//2)+C`. (iv) Put `(2x^(2)+3x+1)=t` so that (4x+3)dx=dt. `:.int((4x+3))/(sqrt(2x^(2)+3x+1))dx=int(dt)/(sqrt(t))=2sqrt(t)+C=2sqrt(2x^(2)+3x+1)+C`. |
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| 46. |
Evaluate : (i) `int((1+cosx))/((1-cosx))dx` (ii) `int((1+sinx))/((1+cosx))dx` |
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Answer» `int((1+cosx))/((a-cosx))dx=int(2cos^(2)(x//2))/(2sin^(2)(x//2))dx` `=intcot^(2)((x)/(2))dx=int("cosec"^(2)(x)/(2)-1)dx` `=int"cosec"^(2)(x)/(2)dx-intdx` `=int"cosec"^(2)tdt-intdx," where"(x)/(2)=t anddx=2dt` `=-2cot t-x+C=-2cot((x)/(2))-x+C`. (ii) `int((1+sinx)/(1+cosx))dx=int(1)/((1+cosx))dx+int(sinx)/((1+cosx))dx` `=int(1)/(2cos^(2)(x//2))dx+int(2sin(x//2)cos(x//2))/(2cos^(2)(x//2))dx` `=(1)/(2)intsec^(2)((x)/(2))dx+int "tan"(x)/(2)dx` `=intsec^(2)tdt+2inttantdt,"where"(x)/(2)=t` `=tant-2log|cost|+C` `=tan((x)/(2))-2log|{:cos((x)/(2)):}|+C`. |
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| 47. |
Evaluate : (i) `int(dx)/(1+sqrt(x))` (ii) `int(x+sqrt(x+1))/(x+2)dx` |
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Answer» (i) Put `sqrt(x)=t` so that `x=t^(2)anddx=2tdt`. `:.int(dx)/(1+sqrt(x))=int(2t)/((1+t))dt=int(2(1+t)-2)/((1+t))dt` `=2intdt-2int(dt)/((1+t))=2t-2log|1+t|+C` `=2sqrt(x)-2log|1+sqrt(x)|+C`. (ii) Put `sqrt(x+1)=t` so that `x+1=t^(2) and dx=2tdt`. `:.int(x+sqrt(x+1))/((x+2))dx=2int((t^(2)-1+t)t)/((t^(2)+1))dt` `=2int((t^(3)+t^(2)-t)/(t^(2)+1))dt` `=2int(t+1-(2t+1)/(t^(2)+1))dt" [by division]"` `=2int(t+1-(2t)/(t^(2)+1)-(1)/(t^(2)+1))dt` `=2inttdt+2intdt-2int(2t)/(t^(2)+1)dt-2int(1)/(t^(2)+1)dt` `=t^(2)+2t-2log|t^(2)+1|-2tan^(-1)t+C` `=(x+1)+2sqrt(x+1)-2log|x+2|-2tan^(-1)sqrt(x+1)+C`. |
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| 48. |
Evaluate `intsqrt((1-x)/(1+x))dx`. |
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Answer» `intsqrt((1+x)/(1-x))dx=int(sqrt(1+x))/(sqrt(1-x))xx(sqrt(1+x))/(sqrt(1+x))dx` `=int(1+x)/(sqrt(1-x^(2)))dx=int(dx)/(sqrt(1-x^(2)))+int(x)/(sqrt(1-x^(2)))dx` `int(dx)/(sqrt(1-x^(2)))-(1)/(2)int(1)/(sqrt(t))dt,"where"(1-x^(2))=t` `=sin^(-1)x-sqrt(t)+C` `=sin^(-1)x-sqrt(t)+C`. |
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| 49. |
Evaluate: `int((3sinx-2)cosx)/(5-cos^2x-4sinx) dx` |
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Answer» We have `I=int((3sinx-2)cosx)/({5-(1-sin^(2)x)-4sinx})dx` `=((3sinx-2)cosx)/({4+sin^(2)x-4sinx})dx=int((3sinx-2)cosx)/((2-sinx)^(2))dx` . . . .(i) Putting 2-sin x=t, we get sin x=2-t and cos x dx =-dt. `:." "I=-int({3(2-t)-2})/(t^(2))dt=-int((4-3t))/(t^(2))dt=int((3t-4))/(t^(2))dt` `=int((3)/(t)-(4)/(t^(2)))dt=3log|t|+(4)/(t)+C` `=3log|{:(2-sinx):}|+(4)/((2-sinx))+C`. |
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| 50. |
Evaluate the following integrals: `inte^(x)cos2xcos4xdx` |
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Answer» Correct Answer - `e^(-x){:[(1)/(74)(6sin6x-cos6x)+(1)/(10)(2sin2x-cos2x)]:}+C` `cos2xcos4x=(1)/(2)(cos6x+cos2x)`. |
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