# Say that integrate from 0 to a of f(t)dt = a^3. Evaluate integrate from 0 to pi/2 of f(cos(x))sin(x)dx

If `int_0^a f(t)dt=a^3` then by derivating `t^3` you get

`f(t)=3t^2`

Then `int_0^(pi/2)f(cos t)sin t dt= int_0^(pi/2)(cos t)^2sin t dt=(1) `

Use substitution `s=cos t` then `ds=-sin t dt` and `s_1= cos 0=1`, `s_2=cos pi/2=0`

`(1)=-int_1^0s^2 ds=1/3`

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If `int_0^a f(t)dt=a^3` then by derivating `t^3` you get

`f(t)=3t^2`

Then `int_0^(pi/2)f(cos t)sin t dt= int_0^(pi/2)(cos t)^2sin t dt=(1) `

Use substitution `s=cos t` then `ds=-sin t dt` and `s_1= cos 0=1`, `s_2=cos pi/2=0`

`(1)=-int_1^0s^2 ds=1/3`

Approved by eNotes Editorial Team

Since `int_0^af(t)dt=a^3` , we want to somehow transform the integral into this form.

`int_0^{pi/2}f(cosx)sinx dx`    let `t=cosx` then `dt=-cosxdx` . limits become t=1 and t=0, so the integral is

`=-int_1^0f(t)dt`  switch the integral limits

`=int_0^1f(t)dt`   now use given information

`=1^3`

`=1`

The integral evaluates to 1.

Approved by eNotes Editorial Team