∫sin3x.cos4x dx
cos7(x)7−cos5(x)5+c
We can see here that the above expression is in ∫sinmx.cosnx dx form with having one power odd and another even. We saw that in such case we substitute the one which has even power.
Since cos(x) has even power we’ll substitute cos (x) = t .
∫sin3x.cos4x dx
Or∫sin2(x).sin(x).cos4(x)dx
Let’s substitute cos(x) = t
- sin(x). dx= dt
So, we’ll have -
−∫(1−t2).t4 dt
Or−∫ t4dt+∫ t6dt
t77−t55+c
Let’s substitute t = cos(x)
And we’ll have
cos7(x)7−cos5(x)5+c