Answer:
We know that, \[2x=(x-\theta )+(x+\theta )\] \[\Rightarrow \] \[\tan 2x=\tan \{(x-\theta )+(x+\theta )\}\] \[\Rightarrow \] \[\tan 2x=\frac{\tan (x-\theta )+(x+\theta )}{1-\tan (x-\theta )\tan (x+\theta )}\] \[\Rightarrow \] \[\tan 2x-\tan (x-\theta )\tan (x+\theta )\tan 2x\] \[=\tan (x-\theta )+\tan (x+\theta )\] \[\Rightarrow \] \[\tan (x-\theta )\tan (x+\theta )\tan 2x=\tan 2x\] \[-\tan (x-\theta )-\tan (x+\theta )\] \[\therefore \] \[l=\int{\tan (x-\theta )\tan (x+\theta )\tan 2x\,dx}\] \[=\int{\{\tan 2x-\tan (x-\theta )-\tan (x+\theta )\}dx}\] \[\Rightarrow \] \[l=-\frac{1}{2}\log |\cos 2x|+\log |\cos (x-\theta )|\] \[+\log |\cos (x+\theta )|+\,C\]
You need to login to perform this action.
You will be redirected in
3 sec
