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
- print Print
- list Cite
Expert Answers
Tibor Pejić
| Certified Educator
calendarEducator since 2012
write609 answers
starTop subjects are Math, Science, and History
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`
Related Questions
- Evaluate the indefinite integral integrate of sin^4(x)cos^4(x)dx
- 1 Educator Answer
- Evaluate the integral : `int_(0)^(pi/2) cos^5 x dx`
- 1 Educator Answer
- Evaluate the integral integrate of (sin(x))^2(cos(x))^4 dx
- 1 Educator Answer
- `int_0^pi sin^2(t) cos^4(t) dt` Evaluate the integral
- 1 Educator Answer
- `int cos(pi/x)/(x^2) dx` Evaluate the indefinite integral.
- 1 Educator Answer
lfryerda
| Certified Educator
calendarEducator since 2012
write738 answers
starTop subjects are Math and Science
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.