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101.

`intlog_10 x dx=`A. `(1)/(x)log_(e)10+C`B. `(1)/(x)log_(10)e+C`C. `x(logx-1)log_(e)10+C`D. `x(logx-1)log_(10)e+C`

Answer» Correct Answer - D
`I=int(logx)/(log10)dx=(1)/(log_(e)10)*underset(I)((logx))*underset(II)(1)dx=(log_(10)e)*int{underset(I)((logx))*underset(II)(1)}dx`
Discussion
102.

`intlogxdx=?`A. `(1)/(x)+C`B. `(1)/(2)(logx)^(2)+C`C. `x(logx+1)+C`D. `x(logx-1)+C`

Answer» Correct Answer - D
`I=intunderset(I)((logx))*underset(II)(1)dx`
Discussion
103.

`intlogxdx=?`

Answer» Integrating by parts, taking log x as the first function and 1 as the second function, we get
`intlogxdx=int(logx*1)dx`
`=(logx)*1dx-int{(d)/(dx)(logx)*int1dx}dx`
`=(logx)*x-int((1)/(x)*x)dx=xlogx-intdx`
`=xlogx-x+C=x(logx-1)+C`.
Discussion
104.

Evaluate the following integrals: `int(x^(2)tan^(-1)x)/((1+x^(2)))dx`

Answer» Correct Answer - `xtan^(-1)x-(1)/(2)log(1+x^(2))-(1)/(2)(tan^(-1)x)^(2)+C`
Put `x=tant,dx=sec^(2)tdtand"then write"tan^(2)t=(sec^(2)t-1)`.
Discussion
105.

Evaluate the following integrals: `int(tanxsec^(2)x)/((1-tan^(2)x))dx`

Answer» Correct Answer - `-(1)/(2)log|1-tan^(2)x|+C`
Put tan x=t and `sec^(2)xdx=dt`.
Discussion
106.

Evaluate: `int(2"x"dot"tan"^(-1)x)/(1+"x"^4)" dx"`A. `(tan^(-1)x^(2))^(2)+C`B. `2tan^(-1)x^(2)+C`C. `(1)/(2)(tan^(-1)x^(2))^(2)+C`D. none of these

Answer» Correct Answer - C
Put `tan^(-1)x^(2)=t and(2x)/((1+x^(4)))dx=dt`.
`:." "I=tdt=(t^(2))/(2)+C=(1)/(2)(tan^(-1)x^(2))^(2)+C`.
Discussion
107.

Evaluate : `int(tan^(-1)x)/((1+x)^(2))dx`.

Answer» Integrating by parts, taking `tan^(-1)`x as the first function and `(1)/((1+x)^(2))` as the second function, we get
`I=tan^(-1)x*((-1))/((1+x))-int(1)/((1+x^(2)))*((-1))/((1+x))dx`
`(-tan^(-1)x)/((1+x))+int(dx)/((1+x)(1+x^(2)))=(-tan^(-1)x)/((1+x))+(1)/(2)*int{(1)/((1-x))+((1-x))/((1+x^(2)))}" [by partial fractions]"`
`=(-tan^(-1))/((1+x))+(1)/(2)log|a+x|+(1)/(2)tan^(-1)x-(1)/(4)log(1+x^(2))+C`.
Discussion
108.

`intsin^(-1)((2x)/(1+x^(2)))dx=?`A. `2xtan^(-1)x+log|1+x^(2)|+C`B. `2xtan^(-1)x-log|1+x^(2)|+C`C. `2xsin^(-1)x+log|1+x^(2)|+C`D. none of these

Answer» Correct Answer - B
Put `x=tant anddx=sec^(2)tdt`.
`:." "I=intsin^(-1)((2tant)/(1+tan^(2)t))sec^(2)tdt=sin^(-1)(sin2t)sec^(2)tdt`
`=2intunderset(I)(t)underset(II)(sec^(2))tdt`.
Discussion
109.

`inttan^(-1)sqrt(x) dx` is equal to

Answer» Put `sqrt(x)=t" so that "(1)/(2sqrt(x))dx=dtordx=2tdt`.
`:.inttan^(-1)sqrt(x)dx=2int(tan^(-1)t)dt`
`=2{:[(tan^(-1)t)*(t^(2))/(2)-int{(1)/((1+t^(2)))*(t^(2))/(2)-int}dt]:}+C`
`=t^(2)(tan^(-1)t)-int(t^(2))/((1+t^(2)))dt+C`
`=t^(2)(tan^(-1)t)-int([(1+t^(2))-1])/((1+t^(2)))dt+C`
`t^(2)(tan^(-1)t)-intdt+int(1)/((1+t^(2)))dt+C`
`=t^(2)(tan^(-1)t)-t+tan^(-1)t+C=(t^(2)+1)tan^(-1)t-t+C`
`=(x+1)tan^(-1)sqrt(x)-sqrt(x)+C`.
Discussion
110.

`int(xtan^-1x)/(1+x^2)^(3/2)dx`

Answer» Put x=tan t so that dx `=sec^(2)tdt`.
`:.int(xtan^(-1)x)/((1+x^(2))^(3//2))dx=int((tant)t)/((1+tan^(2)t)^(3//2))*sec^(2)tdt`
`int((tant)t)/(sect)dt=inttsintdt`
`=t(-cost)-int1*(-cost)-int1*(1-cost)dt" "` [integrating by parts]
`=-tcost+sint+C=(-tan^(-1)x)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+C`
`[becausesint=(x)/(sqrt(1+x^(2)))andcost=(1)/(sqrt(1+x^(2)))]`.
Discussion
111.

`int(xtan^(-1)x^(2))/((1+x^(4)))dx`

Answer» Correct Answer - `(1)/(4)(tan^(-1)x^(2))^(2)+C`
`Put tan^(-1)x^(2)=t`.
Discussion
112.

`int(sin(x-alpha))/(sin(x+alpha))dx=?`A. `xcos2alpha-sin2alpha*log|sin(x+alpha)|+C`B. `xcos2alpha+sin2alpha*log|sin(x+alpha)|+C`C. `xcos2alpha+sinalpha*log|sin(x+alpha)|+C`D. none of these

Answer» Correct Answer - A
`(sin(x-alpha))/(sin(x+alpha))=(sin(x+alpha-2alpha))/(sin(x+alpha))=(sin(x+alpha)cos2alpha-cos(s+alpha)sin2alpha)/(sin(x+alpha))`.
`=cos2alpha-sin2alphacot(x+alpha)`
`:." "I=int{cos2alpha-sin2alpha*cot(x+alpha)}dx`
`=(cos2alpha)*intdx-sin2alpha*intcot(x+alpha)dx`.
`=cos2alpha-sin2alpha*log|sin(x+alpha)|+C`.
Discussion
113.

Evaluate : (i) `intsin^(-1)(3x-4x^(3))dx` (ii) `intsin^(-1)((2x)/(1+x^(2)))dx` (iii) `inttan^(-1)sqrt((1-x)/(1+x))dx` (iv) `intsin^(-1)sqrt((x)/(a+x))dx`

Answer» (i) Put x=sin t so that dx = cos t dt.
`:.intsin^(-1)(2x-4x^(3))dx=intsin^(-1)(3sintsin^(3)t)costdt`
`=intsin^(-1)(3sin3t)costdt`
`=3inttcostdt`
`=3[t(sint)-int*sintdt]" [integrating by parts]"`
`=3tsint+3cost+C`
`=3x(sin^(-1)x)+3sqrt(1-x^(2)+C)`.
(ii) Put x=tan t so that `=sec^(2)tdt`.
`:.intsin^(-1)((2x)/(1+x^(2)))dx=intsin^(-1)((2tant)/(1+tan^(2)t))sec^(2)tdt`
`=intsin^(-1)(sin2t)sec^(2)tdt=2intt*sec^(2)tdt`
`=2t*tant+2log|cost|+C`
`2x(tan^(-1)x)+2log|{:(1)/(sqrt(1+x^(2))):}|+C`
`=2x(tan^(-1)x)+2*(-(1)/(2))log|1+x^(2)|+C`
`=2x(tan^(-1)x)-log|{:1+x^(2):}|+C`.
(iii) Put x=cost so that dx =-sin t dt.
`:,inttan^(-1)sqrt((1-x)/(1+x))dx=inttan^(-1)sqrt((1-cost)/(1+cost))(-sint)dt`
`=inttan^(-1)sqrt((2sin^(2)(t//2))/(2cos^(2)(t//2)))(-sint)dt`
`int{:[tan^(-1)("tan"(t)/(2))]:}(-sint)dt=-(1)/(2)int(sint)dt`
`=-(1)/(2)[t(-cost)-int1*(-cos)dt]" [integrating by parts]"`
`=(1)/(2)t*cos-(1)/(2)sint+C=(1)/(2)x(cos^(-1)x)-(1)/(2)sqrt(1-x^(2))+C`.
(iv) Put `x=atan^(2)t` so that dx `=(2asec^(2)ttant)` dt.
`:.intsin^(-1)sqrt((x)/(a+x))dx=intsin^(-1){sqrt((atan^(2)t)/(a(1+tan^(2)t)))}2asec^(2)t tantdt`
`=2aint t(sec^(2)t*tan t)dt`
`=2a{:[t*(1)/(2)tan^(2)t-int1*(1)/(2)tan^(2)tdt]:}`
[integrating by parts and using `intsec^(2)t tantdt=(1)/(2)tan^(2)t]`
`=at(tan^(2)t)-aint(sec^(2)t-1)dt`
`at(tan^(2)t)-aintsec^(2)t+aintdt`
`at(tan^(2)t)-atant+at+C`
`=xtan^(-1)sqrt((x)/(a))-sqrt(ax)+atan^(-1)sqrt((x)/(a))+C`.
Discussion
114.

`inttan^(-1)sqrt((1-x)/(1+x))dx=`A. `(1)/(2)x(cos^(-1)x)+(1)/(2)sqrt(1-x^(2))+C`B. `(1)/(2)x(sin^(-1)x)+(1)/(2)sqrt(1-x^(2))+C`C. `(1)/(2)x(cos^(-1)x)-(1)/(2)sqrt(1-x^(2))+C`D. none of these

Answer» Correct Answer - C
Put x=cos t and dx =-sin t dt. Then,
`I=inttan^(-1)sqrt((1-cost)/(1+cost))*(-sint)dt=inttan^(-1)sqrt((2sin^(2)((t)/(2)))/(2cos^(2)((t)/(2))))(-sint)dt`
`=-inttan^(-1)("tan"(t)/(2))(-sint)dt=(1)/(2)intunderset(I)(t)underset(II)((sint))dt`.
Discussion
115.

`intx^(2)cosxdx=?`A. `x^(2)sinx+2xcosx-2sinx+C`B. `2xcosx-xsinx+2sinx+C`C. `x^(2)sinx-2xsinx+2sinx+C`D. none of these

Answer» Correct Answer - A
`I=intunderset(I)(x^(2))underset(II)(cos)xdx=x^(2)sinx-intunderset(I)(2x)underset(II)(sin)xdx`
`=(x^(2)sinx)-2[x(-cosx)-int(-cosx)dx]+C`
`=(x^(2)sinx)+2xcosx-2sinx+C`.
Discussion
116.

`int(x^4)/(1+x^2)dx`A. `(x^(3))/(3)+x+tan^(-1)x+C`B. `(-x^(3))/(3)+x-tan^(-1)x+C`C. `(-x^(3))/(3)-x+tan^(-1)x+C`D. none of these

Answer» Correct Answer - C
On dividing `x^(4)` by `(x^(2)+1)`, we get :
`I=int{(x^(2)-1)+(1)/(x^(2)+1)}dx=(x^(3))/(3)-x+tan^(-1)x+C`.
Discussion
117.

`int(xtan^-1x)/(1+x^2)^(3/2)dx`A. `(tan^(-1)x)/(sqrt(1+x^(2)))-(x)/(sqrt(1+x^(2)))+C`B. `(-tan^(-1)x)/(sqrt(1+x^(2)))+(x)/(sqrt(1+x^(2)))+C`C. `(xtan^(-1)x)/(sqrt(1+x^(2)))+(1)/(2)log|(x)/(sqrt(1+x^(2)))|+C`D. none of these

Answer» Correct Answer - B
Put x=tan t, so that dx `=sec^(2)tdtandt=tan^(-1)x`.
`:." "I=int(t(tant))/((1+tan^(2)t)^((3)/(2)))sec^(2)tdt=intunderset(I)(t)underset(II)(sin)tdt`.
Discussion
118.

`inttan^(-1)sqrt(x) dx` is equal toA. `(x-1)tan^(-1)sqrt(x)+sqrt(x)+C`B. `(x+1)tan^(-1)sqrt(x)-sqrt(x)+C`C. `(1)/(2)sqrt(x)tan^(-1)sqrt(x)-(1)/(2)sqrt(x)+C`D. none of these

Answer» Correct Answer - B
Put `sqrt(x )=t and(1)/(2sqrt(x))dx=dtordx=2tdt`.
`:." "I=2intunderset(II)(t)underset(I)((tan^(-1)t))dt`.
Discussion
119.

`intcos^(-1)((1-x^(2))/(1+x^(2)))dx=?`A. `2xtan^(-1)x+log(1+x^(2))+C`B. `-2xtan^(-1)x-2log(1+x^(2))+C`C. `2xtan^(-1)x-log(1+x^(2))+C`D. none of these

Answer» Correct Answer - C
Put `x=tant anddx=sec^(2)tdt`.
Then, `I=intcos^(-1)((1-tan^(2)t)/(1+tan^(2)t))sec^(2)tdt=intcos^(-1)(cos2t)sec^(2)tdt`
`=2intunderset(I)(t)underset(II)(sec^(2))tdt`.
Discussion
120.

Evaluate `int sin(logx)dx`.A. `(1)/(2)xsinlogx+(1)/(2)xcos(logx)+C`B. `(1)/(2)xsinlogx-(1)/(2)xcos(logx)+C`C. `-(1)/(2)xsin(logx)+(1)/(2)xcos(logx)+C`D. none of these

Answer» Correct Answer - B
`I=int{sinunderset(I)((logx))*underset(II)1}dx=(sinlogx)x-int(cos(logx))/(x)*xdx`
`=xsin(logx)-int{cosunderset(I)((logx))*underset(II)(1)}dx`
`=xsin(logx)-cos(logx)*x+int(-sin(logx))/(x)*xdx`
`:." "2I=xsin(logx)-xcos(logx)-C`
Discussion
121.

`intxsin^(3)x^(2)cosx^(2)dx=?`A. `(1)/(4)sin^(4)x^(2)+C`B. `(1)/(8)sin^(4)x^(2)+C`C. `(1)/(2)sin^(4)x^(2)+C`D. none of these

Answer» Correct Answer - B
Put `sinx^(2)=t and2xcosx^(2)dx=dt`.
`:." "I=(1)/(2)intt^(3)dt=(1)/(2)xx(t^(4))/(4)+C=(t^(4))/(8)+C=(1)/(8)sin^(4)x^(2)+C`.
Discussion
122.

Evaluate: `inttan^(-1)((3x-x^3)/(1-3x^2)) dx`A. `3xtan^(-1)x+(3)/(2)log(1+x^(2))+C`B. `3xtan^(-1)x-(3)/(2)log(1+x^(2))+C`C. `3xcos^(-1)x-(3)/(2)sqrt(1+x^(2))+C`D. `3xsin^(-1)x+(3)/(2)sqrt(1+x^(2))+C`

Answer» Correct Answer - B
Put `x=tant andtdx=sec^(2)tdt`.
`I=inttan^(-1)(tan3t)sec^(2)tdt=3underset(I)(t)underset(II)(sec^(2))tdt`.
Discussion
123.

`int ((1+ sinx))/((1- sinx)) dx=?`A. `2tanx+x-2secx+C`B. `2tanx-x+2secx+C`C. `2tanx-x-2secx+C`D. none of these

Answer» Correct Answer - B
`I=int((1+sinx))/((1-sinx))xx((1+sinx))/((a+sinx))dx`.
`=int((1+sinx)^(2))/((1-sin^(2)x))dx=int(1+sin^(2)x+2sinx)/(cos^(2)x)dx`
`=int{(1)/(cos^(2)x)+(sin^(2)x)/(cos^(2)x)+(2sinx)/(cos^(2)x)}dx`
`int(sec^(2)x+tan^(2)x+2secxtanx)dx`
`int(2sec^(2)x-1+2secxtanx)dx=tanx-x+2secx+C`.
Discussion
124.

`intx.tan^-1xdx`A. `(1)/(2)(x^(2)+1)tan^(-1)x-(1)/(2)x+C`B. `(1)/(2)(x^(2)-1)tan^(-1)x-(1)/(2)x+C`C. `(1)/(2)(x^(2)+1)tan^(-1)x+(1)/(2)x+C`D. none of these

Answer» Correct Answer - A
`I=intunderset(II)(x)underset(I)(tan^(-1))xdx=(tan^(-1)x)*(x^(2))/(2)-(1)/((1+x^(2)))*(x^(2))/(2)dx`
`=(1)/(2)x^(2)(tan^(-1))-(1)/(2)int(1-(1)/(1+x^(2)))dx=(1)/(2)x^(2)tan^(-1)x-(1)/(2)x+(1)/(2)tan^(-1)x+C`
`=(1)/(2)(x^(2)+1)tan^(-1)x-(1)/(2)x+C`
Discussion
125.

`intx.tan^-1xdx`A. `(1)/(2)tan^(-1)x+log(1+x^(2))-(1)/(2)x+C`B. `(1)/(2)x^(2)tan^(-1)x+(1)/(2)+C`C. `(1)/(2)(1+x)^(2)tan^(-1)x-(1)/(2)x+C`D. none of these

Answer» Correct Answer - C
`I=intunderset(II)(x)underset(I)((tan^(-1)x))dx`.
Discussion
126.

`inte^(cos^(2))sin2xdx=?`A. `e^(cos^(2)x)+C`B. `-e^(cos^(2)x)+C`C. `e^(sin^(2)x)+C`D. none of these

Answer» Correct Answer - B
Put `cos^(2)x=t and-2cosxsinxdx=dt`.
`:." "I=-inte^(t)dt=-e^(t)+C=-e^(cos^(2))+C`.
Discussion
127.

Evaluate : (i) `intcos2x dx` (ii) `inte^((5x+3))dx` (iii) `int sec^(2)(3x+5)dx` (iv) `intsin^(3)xdx`

Answer» (i) Put `2x="t so that "2dx=dt" or "=(1)/(2)dt`.
`:." "intcos2x=(1)/(2)intcosdt=(1)/(2)sint+C=(1)/(2)sin2x+C`.
(ii) Put (5x+3)=t so that 5dx=dt or dx `=(1)/(5)dt`
`:." "inte^((5x+3))dx=(1)/(5)*e^(t)+C=(1)/(5)e^((5x+3))+C`.
(iii) Put (3x+5)dx=t so that 3 dx = dt or `dx=(1)/(3)dt`.
`:.intsec^(2)(3x+5)dx=(1)/(3)=intsec^(2)t" dt"=(1)/(3)tant+C=(1)/(3)tan(3x+5)+C`.
(iv) We know that `sin3x=3 sin x-4sin^(3)x`.
`:." "sin^(3)x=(1)/(4)(3sinx-sin3x)`.
so, `intsin^(3)xdx=int((3)/(4)sinx-(1)/(4)sin3x)dx=(3)/(4)intsinxdx-(1)/(4)fsin3xdx=(3)/(4)intxdx-(1)/(4)(-cosx)-(1)/(4)*(-cos3x)/(3)+C=-(3)/(4)cosx+(cos3x)/(12)+C`.
Discussion
128.

`inttan^(-1)(secx+tanx)dx=?`A. `(pix)/(4)+(x^(2))/(4)+C`B. `(pix)/(4)-(x^(2))/(4)+C`C. `(1)/((1+x^(2)))+C`D. none of these

Answer» Correct Answer - A
`tan^(-1)(secx+tanx)=tan^(-1)((1)/(cosx)+(sinx)/(cosx))=tan^(-1){(1+sinx)/(cosx)}`.
`=tan^(-1){(1-cos((pi)/(2)+x))/(sin((pi)/(2)+x))}=tan^(-1){(2sin^(2)((pi)/(4)+(x)/(2)))/(2sin((pi)/(4)+(x)/(2))cos((pi)/(4)+(x)/(2)))}`
`=tan^(-1){tan((pi)/(4)+(x)/(2))}=((pi)/(4)+(x)/(2))`
`:." "I=int((pi)/(4)+(x)/(2))dx(pix)/(4)+(x^(2))/(4)+C`.
Discussion
129.

`intsecxlog(secx+tanx)dx`

Answer» Correct Answer - `(1)/(2)[log(secx+tanx)]^(2)+C`
`Putlog(secx+tanx)=t`.
Discussion
130.

`int(3-5x)^(7)dx=?`A. `-5(3-5x)^(6)+C`B. `((3-5x)^(8))/(-40)+C`C. `(-5(3-5x)^(8))/(8)+C`D. none of these

Answer» Correct Answer - B
Put(3-5x)=t and -5dx=dt.
Discussion
131.

`intsqrt(1+x)dx`

Answer» Correct Answer - `(2)/(3)(1+x)^(3//2)+C`
Discussion
132.

`intsec^(4)xtanxdx=?`A. `(1)/(2)sec^(2)x+(1)/(4)sec^(4)x+C`B. `(1)/(2)tan^(2)x+(1)/(4)tan^(4)x+C`C. `(1)/(2)secx+log|secx+tanx|+C`D. none of these

Answer» Correct Answer - B
`I=intsec^(2)x*sec^(2)x*tanxdx`.
`=intsec^(2)x(1+tan^(2)x)tanxdx=int(1+t^(2))tdt`, where tan x=t
`int(t+t^(3))dt=(t^(2))/(2)+(t^(4))/(4)+C=(1)/(2)tan^(2)x+(1)/(4)tan^(4)x+C`.
Discussion
133.

`intsec^-1xdx`A. `xsec^(-1)x+log|x+sqrt(x^(2)-1)|+C`B. `xsec^(-1)x-log|x+sqrt(x^(2)-1)|+C`C. `xsec^(-1)x+log|x-sqrt(x^(2)-1)|+C`D. none of these

Answer» Correct Answer - B
Put `sec^(-1)x=t," so that "x=sect anddx=sect tantdt`.
`:." "I=intunderset(I)(t)underset(II)((sect tant))dt`.
Discussion
134.

`int(cosx)/((1+cosx))dx=?`A. `x+"tan"(x)/(2)+C`B. `-x+"tan"(x)/(2)+C`C. `x-"tan"(x)/(2)+C`D. none of these

Answer» Correct Answer - C
`I=int{((1+cos)-1)/((1+cosx))}=f{1-(1)/(1+cosx)}dx`.
`intdx-int(dx)/(1+cosx)=x-int(dx)/(2cos^(2)((x)/(2)))=-(1)/(2)int"sec"^(2)(x)/(2)dx`
`=x-(1)/(2)xx2xxintsec^(2)"dt,where"(x)/(2)=t`
`=x-tantC=x-"tan"(x)/(2)+C`.
Discussion
135.

Evaluate the following integrals :`int(1-cos2x)/(1+cos2x)dx`A. `tanx+x+C`B. `tanx-x+C`C. `-tanx+x+C`D. none of these

Answer» Correct Answer - B
`I=int(2sin^(2)x)/(2cos^(2)x)dx=inttan^(2)xdx=int(sec^(2)x-1)dx=tanx-x+C`.
Discussion
136.

`inte^(cos^(2))sin2xdx=?`

Answer» Correct Answer - `-e^(cos^(2)x)+C`
Put `cos^(2)x=t`.
Discussion
137.

`int(cotx)/(log(sinx))dx`A. `log|cotx|+C`B. `log|cotx"cosec "x|+C`C. `log|logsinx|+C`D. none of these

Answer» Correct Answer - C
Put log (sin x) = t and cot x dx = dt.
Discussion
138.

`int(dx)/((3-5x))`

Answer» Correct Answer - `(-log|3-5x|)/(5)+C`
Discussion
139.

Evaluate : (i) `int(1)/((1+tanx))dx` (ii) `int(1)/((1+cotx))dx` (iii) `int((1-tanx)/(1+tanx))dx` (iv) `int(tanx)/((secx+cosx))dx`

Answer» (i) `int(1)/((1+tanx))dx=int(1)/((1+(sinx)/(cosx)))dx`
`=int(cosx)/((cosx+sinx))dx=int((cosx+sinx)+(cosx-sinx))/(2(cosx+sinx))dx`
`=(1)/(2)intdx+(1)/(2)int((cosx-sinx))/((cosx+sinx))dx`
`=(1)/(2)intdx+(1)/(2)int(1)/(t)dt," where"(cosx+sinx)=tand(cosx-sinx)dx=dt`
`=(1)/(2)x+(1)/(2)log|t|+C=(1)/(2)x+(1)/(2)log|{:cosx+sinx:}|+C`
(ii) `int(1)/((1+cotx))dx=int(1)/((1+(cosx)/(sinx)))dx=int(sinx)/((sinx+cosx))dx`
`=int((sinx+cosx)-(cosx-sinx))/(2(sinx+cosx))dx`
`=(1)/(2)intdx-(1)/(2)int((cosx-sinx))/((sinx+cosx))dx`
`=(1)/(2)intdx-(1)/(2)int(1)/(t)dt`,
where sin x + cos x = t and (cos x - sin x)dx=dt
`=(1)/(2)x-(1)/(2)log|t|+C=(1)/(2)x-(1)/(2)log|{:sinx+cosx:}|+C`.
(iii) `int((1-tanx)/(1+tanx))dx=int((1-(sinx)/(cosx)))/((1+(sinx)/(cosx)))dx=int((cosx-sinx))/((cosx+sinx))dx`
`=int(1)/(t)dt," where"(cosx+sinx)=t and(cosx+sinx)dx=dt`
`=log|t|+C=log|{:(cosx+sinx):}|+C`.
(iv) `int(tanx)/((secx+cosx))dx=int(sinx)/(1+cos^(2)x)dx`
`=-int(1)/((1+t^(2)))dt," where"cosx=t and sinx dx=-dt`,
`=-tan^(-1)t+C=-tan^(-1)(cosx)+C`.
Discussion
140.

`int (dx)/(e^x + e^-x)`

Answer» Correct Answer - `tan^(-1)(e^(x))+C`
Discussion
141.

`int(sec^(2)x)/((1+tanx))dx`

Answer» Correct Answer - `log|1+tanx|+C`
Discussion
142.

Evaluate `intsec^(2)x"cosec"^(2)xdx`.A. `tanx-cotx+C`B. `tanx+cotx+C`C. `-tanx+cotx+C`D. none of these

Answer» Correct Answer - A
`I=intsec^(2)x(1+cot^(2))dx=intsec^(2)xdx+intsec^(2)xcot^(2)xdx`.
`=tanx+int(1)/(cos^(2)x)*(cos^(2)x)/(sin^(2)x)dx=tanx+int"cosec"^(2)xdx`
`=tanx-cotx+C`.
Discussion
143.

`intsec^(3)xdx=?`A. `(1)/(2){secxtanx-log|secx+tanx|}+C`B. `(1)/(2){secxtanx+log|secx+tanx|}+C`C. `2{secxtanx+log|secx+tanx|}+C`D. none of these

Answer» Correct Answer - B
`I=intunderset(II)(sec^(2))x*underset(I)(sec)xdx=secxtanx-intsecxtan^(2)xdx`
`=secxtanx-intsecx(sec^(2)x-1)dx=secxtanx-I+int+secxdx`
`:." "2I=secxtanx+log|secx+tanx|+C`.
Discussion
144.

Integrate `sin(ax+b)cos(ax+b)`

Answer» Correct Answer - `(sin^(2)(ax+b))/(2a)+C`
Put sin (ax+b)=t.
Discussion
145.

`int 1/x^2 e^(-1/x) dx=`

Answer» Correct Answer - `e^(-1//x)+C`
Put `(-1)/(x)=t`.
Discussion
146.

`int(dx)/((1-cotx))=?`

Answer» Correct Answer - `(1)/(2)x+(1)/(2)log|sinx-cosx|+C`
Discussion
147.

Evaluate : (i) `int cos^(3)xsinxdx` (ii) `int(sqrt(sinx))cosxdx` (iii) `int(cosec^(2)x)/((1+cotx))dx` (iv) `int(sinx)/((3+4cosx)^(2))dx`

Answer» (i) Put cos x=t so that sin x dx =-dt.
`:.intcos^(3)xsinxdx=-intt^(3)dt=-(t^(4))/(4)+C=-(1)/(4)cos^(4)x+C`.
(ii) Put sin x=t so that cos x dx = dt.
`:.int(sqrt(sinx))cosxdx=intsqrt(t)dt=(2)/(3)t^(3//2)+C=(2)/(3)(sinx)^(3//2)+C`.
(iii) Put (1+cot x)=t so that `-cosec^(2)xdx=dt`.
`:.int("cosec"^(2)x)/((1+cotx))dx=-int(1)/(t)dt=-logt+C=-log|{:(1+cotx):}|+C`.
(iv) Put (3+4cos x) =t so that -4sin x dx=dt.
`:.int(sinx)/((3+4cosx)^(2))dx=-(1)/(4)int(1)/(t^(2))dt=(1)/(4t)+C=(1)/(4(3+4cosx))+C`.
Discussion
148.

`inte^(sinx)sin2x dx`A. `(2sinx)e^(sinx)+C`B. `(2cosx)e^(sinx)+C`C. `2e^(sinx)(sinx+1)+C`D. `2e^(sinx)(sinx-1)+C`

Answer» Correct Answer - D
`I=2inte^(sinx)sinxcosxdx=2intunderset(I)(t)underset(II)e^(t)"dt, where "sinx=t`.
Discussion
149.

`int(cotx)/(log(sinx))dx`

Answer» Correct Answer - `log|log(sinx)|+C`
Discussion
150.

Evaluate the following integrals: `int((1-cotx)/(1+cotx))dx`

Answer» Correct Answer - `-log|sinx+cosx|+C`
`I=-int((cosx-sinx))/((sinx+cosx))dx.Put(sinx+cosx)=t`.
Discussion