Question

i need the solution for 134 question

Answer 1

Let I=∫sec2(x)(sec(x)+tan(x))5dxI=∫sec2⁡(x)(sec⁡(x)+tan⁡(x))5dx

Lets apply integration by parts technique,

Assume u=sec(x)⟹du=sec(x)tan(x)dxu=sec⁡(x)⟹du=sec⁡(x)tan⁡(x)dx

and dv=sec(x)(sec(x)+tan(x))5dxdv=sec⁡(x)(sec⁡(x)+tan⁡(x))5dx

⟹v=∫sec2(x)(sec(x)+tan(x))5dx⟹v=∫sec2⁡(x)(sec⁡(x)+tan⁡(x))5dx

=∫sec(x)(sec(x)+tan(x))(sec(x)+tan(x))6dx=∫sec⁡(x)(sec⁡(x)+tan⁡(x))(sec⁡(x)+tan⁡(x))6dx

=∫1(sec(x)+tan(x))6d(sec(x)+tan(x))=∫1(sec⁡(x)+tan⁡(x))6d(sec⁡(x)+tan⁡(x))

=−5(sec(x)+tan(x))5=−5(sec⁡(x)+tan⁡(x))5

As, ∫udv=uv−∫vdu∫udv=uv−∫vdu

So, I=−5sec(x)(sec(x)+tan(x))5+∫5sec(x)tan(x)(sec(x)+tan(x))5dxI=−5sec⁡(x)(sec⁡(x)+tan⁡(x))5+∫5sec⁡(x)tan⁡(x)(sec⁡(x)+tan⁡(x))5dx

=−5sec(x)(sec(x)+tan(x))5+∫5sec(x)tan(x)+5sec2(x)−5sec2(x)(sec(x)+tan(x))5dx=−5sec⁡(x)(sec⁡(x)+tan⁡(x))5+∫5sec⁡(x)tan⁡(x)+5sec2⁡(x)−5sec2⁡(x)(sec⁡(x)+tan⁡(x))5dx

=−5sec(x)(sec(x)+tan(x))5+∫5sec(x)tan(x)+5sec2(x)(sec(x)+tan(x))5dx−∫5sec2(x)(sec(x)+tan(x))5dx=−5sec⁡(x)(sec⁡(x)+tan⁡(x))5+∫5sec⁡(x)tan⁡(x)+5sec2⁡(x)(sec⁡(x)+tan⁡(x))5dx−∫5sec2⁡(x)(sec⁡(x)+tan⁡(x))5dx

⟹I=−5sec(x)(sec(x)+tan(x))5+∫5sec(x)(tan(x)+sec(x))(sec(x)+tan(x))5dx−5I⟹I=−5sec⁡(x)(sec⁡(x)+tan⁡(x))5+∫5sec⁡(x)(tan⁡(x)+sec⁡(x))(sec⁡(x)+tan⁡(x))5dx−5I

⟹6I=−5sec(x)(sec(x)+tan(x))5−20(sec(x)+tan(x))4⟹6I=−5sec⁡(x)(sec⁡(x)+tan⁡(x))5−20(sec⁡(x)+tan⁡(x))4

⟹I=−25sec(x)+20tan(x)6(sec(x)+tan(x))5+C