InterviewSolution
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InterviewSolution
| 1. |
Evaluate `int(sqrt(tanx)+sqrt(cotx))dx`. |
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Answer» We have `int(sqrt(tanx)+sqrt(cotx))dx` `=int(sqrt(tanx)+(1)/(sqrt(tanx)))dx=int((tanx+1))/(sqrt(tanx))dx` `=int((t^(2)+1))/(t)*(2t)/((1+t^(4)))dt " where tan" =t^(2)rArrx=tan^(-1)t^(2)rArrdx=(2t)/((1+t^(4)))dt` `=2int((t^(2)+1))/((t^(4)+1))dt=2int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt` [ on dividing num . and denom by `t^(2)`] `=2int((1+(1)/(t^(2))))/((t-(1)/(t))^(2)+2)dt` `=2int(du)/((u^(2)+2))`, where ` (t-(1)/(t))=u and (1+(1)/(t^(2)))dt=du` `=2*(1)/(sqrt(2))"tan"^(-1)(u)/(sqrt(2))+C=sqrt(2)"tan"^(-1)((t-(1)/(t)))/(sqrt(2))+C` `[becauseu=(t-(1)/(t))]` `=sqrt(2)"tan"^(-1)((t^(2)-1))/((sqrt(2)t))+C=sqrt(2)tan^(-1)((tanx-1)/(sqrt(2tanx)))+C [ becauset^(2)=tanx]`. |
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