Q.

The integral ∫0π4136sinx3sinx+5cosx dx is equal to:            [2024]

1 3π-30loge2+20loge5  
2 3π-50loge2+20loge5  
3 3π-25loge2+10loge5  
4 3π-10loge(22)+10loge5  

Ans.

(2)

Let I=∫0π/4136sinx3sinx+5cosx dx

Let sinx=λ(3sinx+5cosx)+μ(3cosx-5sinx)

For x=0, we have 0=5λ+3μ

and for x=π/2, we have 1=3λ-5μ

Solving these two equations, we get λ=334 and μ=-534

∴     I=136∫0π/4334(3sinx+5cosx3sinx+5cosx)dx-136×534∫0π/43cosx-5sinx3sinx+5cosx dx

=136×334[x]0π/4-136×534ln|3sinx+5cosx|0π/4

=3π-20[ln82-ln5]=3π-20ln42+20ln5

=3π-20ln(2)52+20ln5=3π-50ln2+20ln5