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

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

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Answer 1 · 2016-09-25T10:09:35

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

Let #sqrtx=y rArr x=y^2 rArr dx=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.!