Question

Integral (sin t dt)/(25 + cos^2 t)?

Answer 1

`=-1/5tan^-1(cost/5)+c`

Explanation:

Please check this site: https://www.wolframalpha.com/input/?i=integral+sintdt%2F(25%2Bcos%5E2t)
Substitute `1/5cost` as x.
`1/5(-sintdt) = dx`

Thus the integral becomes:
`int(-5)/(25(1+x^2))dx`

`=-1/5tan^-1(x)`
substituting x back, we get, `=-1/5tan^-1(cost/5)+c`