How do you find int5/(16+9cos^2x)dx516+9cos2xdx?

2 Answers
Jul 22, 2018

I=1/4arc tan (4/5tanx)+cI=14arctan(45tanx)+c

Explanation:

Here ,

I=int5/(16+9cos^2x)dxI=516+9cos2xdx

=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

Jul 22, 2018

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.