InterviewSolution
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InterviewSolution
| 1. |
`int 1/(sinx+sin2x)dx`. |
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Answer» `I = int 1/(sinx+sin2x)dx` `= int 1/(sinx+2sinxcosx)dx` `= int 1/(sinx(1+2cosx)dx` `= int sinx/(sin^2x(1+2cosx)dx` `= int sinx/((1-cos^2x)(1+2cosx)dx` Let `cosx = t , then -sinxdx= dt` `:. I = int -dt/((1-t^2)(1+2t) )` `= int -dt/((1-t)(1+t)(1+2t) )` Now, `1/((1-t)(1+t)(1+2t)) = A/(1-t)+B/(1+t)+C/(1+2t)` `=> 1/((1-t)(1+t)(1+2t)) = (A(1+t)(1+2t)+B(1-t)(1+2t)+C(1-t)(1+t))/((1-t)(1+t)(1+2t))` `=> 1/((1-t)(1+t)(1+2t)) = (A+3At+2At^2+B+2Bt-2Bt^2+C-Ct^2)/((1-t)(1+t)(1+2t))` Now, comparing coefficients on both sides, `0 = 2A-2B - C->(1)` `0 = 3A+3B->(2)` `1 = A+B+C->(3)` Solving (1),(2), and (3), we get, `A = 1/6, B = -1/2, C = 4/3` So, our integral becomes, `I =- int 1/(6(1-t))-1/(2(1+t))+4/(3(1+2t)) dt` `I =- int 1/(6(1-t)) + int 1/(2(1+t)) - int 4/(3(1+2t)) dt` `I = -1/6 ln(1-t)(-1) +1/2 ln(1+t) - 4/3ln(1+2t)(1/2) +c` `I = 1/6 ln(1-cosx) +1/2 ln(1+cosx) - 2/3ln(1+2cosx) +c` |
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