How do you integrate int sec^2sqrtxsec2x by integration by parts method?

How do you integrate int sec^2sqrtxdxsec2xdx by integration by parts method?

1 Answer
Sep 25, 2016

2[sqrtxtan(sqrtx)-ln|tansqrtx+secsqrtx|]+C2[xtan(x)lntanx+secx]+C.

Explanation:

Let sqrtx=y rArr x=y^2 rArr dx=2ydyx=yx=y2dx=2ydy.

:. I=intsec^2sqrtxdx=int(sec^2y)(2y)dy=2intysec^2ydy.

The Rule of Integration by Parts (IbP) states :

" (IbP) : "intuvdy=uintvdy-int((du)/dy*intvdy)dy.

We take, u=y, and, v=sec^2y.

:. (du)/dy=1, and, intvdy=tany.

:. I=2[ytany-inttanydy]

=2[ytany-ln|tany+secy|}

Replacing y" by "sqrtx, we have,

I=2[sqrtxtan(sqrtx)-ln|tansqrtx+secsqrtx|]+C.

Enjoy maths.!