`int_0^(pi/2) sin^7(theta) cos^5(theta) d theta` Evaluate the integral

`int_0^(pi/2)sin^7(theta)cos^5(theta)d theta`

Let's first compute the indefinite integral by rewriting the integrand,

`intsin^7(theta)cos^5(theta)d theta=intsin^6(theta)sin(theta)cos^5(theta)d theta`

Now use the identity:`sin^2(x)=1-cos^2(x)`

`=int(1-cos^2(theta))^3sin(theta)cos^5(theta)d theta`

Now apply integral substitution,

Let `u=cos(theta)`

`=>du=-sin(theta)d theta`

`=int(1-u^2)^3u^5(-du)`

`=-int(1-u^2)^3u^5du`

`=-int(1-u^6-3u^2+3u^4)u^5du`

`=-int(u^5-u^11-3u^7+3u^9)du`

`=-intu^5du+intu^11du+3intu^7du-3intu^9du`

`=-u^6/6+u^12/12+3(u^8/8)-3(u^10/10)`

`=u^12/12-3/10u^10+3/8u^8-1/6u^6`

Substitute back `u=cos(x)`

`=1/12cos^12(x)-3/10cos^10(x)+3/8cos^8(x)-1/6cos^6(x)`

Add a constant C to the solution,

`=1/12cos^12(x)-3/10cos^10(x)+3/8cos^8(x)-1/6cos^6(x)+C`

Now let's evaluate the definite integral,

`int_0^(pi/2)sin^7(theta)cos^5(theta)d theta=[1/12cos^12(x)-3/10cos^10(x)+3/8cos^8(x)-1/6cos^6(x)]_0^(pi/2)`

`=[1/12cos^12(pi/2)-3/10cos^10(pi/2)+3/8cos^8(pi/2)-1/6cos^6(x)]-[1/12cos^12(0)-3/10cos^10(0)+3/8cos^8(0)-1/6cos^6(0)]`

Plug in the values of `cos(pi/2)=0.cos(0)=1`

`=[0]-[1/12-3/10+3/8-1/6]`

`=-[(10-36+45-20)/120]`

`=-[-1/120]`

`=1/120`