Question

How do you find `int5/(16+9cos^2x)dx`?

Answer 1

`I=1/4arc tan (4/5tanx)+c`

Explanation:

Here ,

`I=int5/(16+9cos^2x)dx`

#=int(5sec^2x)/(16sec^2x+9)dx.........to[becausecostheta=1/sectheta]#

`=int(5sec^2x)/(16(1+tan^2x)+9)dxto[becausesec^2 theta-tan^2theta=1]`

`=5intsec^2x/(16tan^2x+25)dx`

Subst. `color(blue)(tanx=u=>sec^2xdx=du`

`:.I=5int1/(16u^2+25)du`

`=5/16int1/(u^2+25/16)du`

`=5/16int1/(u^2+(5/4)^2)du`

`=5/16*1/(5/4)arctan(u/(5/4))+c`

`=5/16 xx 4/5arc tan((4u)/5)+c`

`:.I=1/4arctan((4u)/5)+c`

Subst. back ,`color(blue)(u=tanx`

`I=1/4arc tan (4/5tanx)+c`

Answer 2

`1/4arctan(4/5tanx)+C`.

Explanation:

Let, `I=int5/(16+9cos^2x)dx`.

`:. I=5int1/{cos^2x(16/cos^2x+9)dx`,

`=5int{(1/cos^2x)(1/(16sec^2x+9))}dx`,

`=5int{1/{16(tan^2x+1)+9}}sec^2xdx`.

This suggests that the substn. `4tanx=u` must work.

Now, `4tanx=u rArr sec^2xdx=1/4du`.

`:. I=5int{1/(u^2+5^2)}1/4du`,

`=5/4*1/5arctan(u/5)`.

`rArr I=1/4arctan(4/5tanx)+C`.