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

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Answer 1 · 2018-07-22T14:28:20

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

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 · 2018-07-22T14:43:16

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

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#.