Question

integration of xsec^2x.tanx dx

Answer 1

Answer:

The answer is explained above.

Answer 2

To find : $\int x\:sec^{2}x\:tanx\:dx$ = ?

Solution :

Let, tanx = z

⇒ sec²x dx = dz

and z = tan⁻¹x

Now, $\int x\:sec^{2}x\:tanx\:dx$

$=\int z\:tan^{-1}z\:dz$

$=tan^{-1}z \int z\:dz-\int\{\frac{d}{dz}(tan^{-1}z)\times \int z\:dz\}dz$

$=\frac{z^{2}}{2}\:tan^{-1}z-\frac{1}{2}\int \frac{z^{2}}{z^{2}+1}dz$

$=\frac{z^{2}}{2}\:tan^{-1}z-\frac{1}{2}\int \big(1-\frac{1}{z^{2}+1}\big)dz$

$=\frac{z^{2}}{2}\:tan^{-1}z-\frac{z}{2} +\frac{1}{2}tan^{-1}z+C$ ,

where C is constant of integration

$=\frac{x\:tan^{2}x}{2}-\frac{tanx}{2}+\frac{x}{2}+C$ ,

which is the required integral. (Ans.)