∫π/40(sinx+cosx)9+16sin2xdx
I=∫π40sinx+cosx9+16sin2xdx
Let sinx−cosx=t
(cosx+sinx)dx=dt
Again (sinx−cosx)2=t2
⇒sin2x+cos2x−2sinxcosx=t2
⇒1−sin2x=t2
⇒sin2x=1−t2
When x=0⇒t=sin0−cos0=−1
When x=π4⇒t=sinπ4−cosπ4=1√2−1√2=0
∴I=∫0−1dt9+16(1−t2)
=∫0−1dt9+16−16t2
=∫0−1dt25−16t2
=116∫0−1dt2516−t2
=116⎡⎢
⎢
⎢⎣12×54log∣∣
∣
∣
∣∣54+t54−t∣∣
∣
∣
∣∣⎤⎥
⎥
⎥⎦0−1
=14[110log∣∣∣5+4t5−4t∣∣∣]0−1
=140[log∣∣∣5+05−0∣∣∣−log∣∣∣5−45+4∣∣∣]
=140[log1−log(19)]
=140[0+log9]
=log940