∫sin3(x).cos5(x)dx
cos8(x)8−cos6(x)6+C
We can see that this is the Integration of type ∫sinmx.cosnx dx. Out of all the cases discussed, we can see that it is the case where m & n both are odd. So we can substitute either sin(x) or cos(x) equal to t.
Let’s substitute cos(x) = t
So, - sin(x) dx = dt
∫sin3(x).cos5(x)dx=∫sin2(x).cos5(x).sin(x)dxOr −∫(1−t2).t5.dt∫(t2−1).t5.dt∫(t7−t5).dt=t88−t66+c
Substitute t = cosx
=cos8(x)8−cos6(x)6+c