check out Define the integral as I convert sin(x)=cos(pi/2-x) Apply kings property cos(pi/2-(2pi-x))=cos(-3pi/2+x) cos(x-3pi/2) Use the sum/difference Add two forms of I Play around using algebra
Split interval of integration into quadrants Integral on first quadrant leave as it is In integral on second quadrant use substitution t = \pi-x In integral on third quadrant use substitution t = x - pi In integral on fourth quadrant use substitution t = 2\pi - x Add these integrals to get integral 4\int_{0}^{\pi}{\frac{25+9sin^2(x)}{(25-9sin^2(x))^2}dx} Use substitution t = tan(x) to get 4\int_{0}^{\infty}{\frac{25+34t^2}{(25+16t^2)^2}}dt} Now we can rewrite it as 4(\int_{0}^{\infty}{\frac{25+16t^2}{(25+16t^2)^2}}dt} + 18\int_{0}^{\infty}{\frac{t^2}{(25+16t^2)^2}dt}) Then integrate \int_{0}^{\infty}{\frac{t^2}{(25+16t^2)^2}dt} by parts
Another potential solution: Consider I(t)=integral 0 to 2π of 1/(t-3sinx). Calculate I(t) using standard techniques or Weierstrass substitution. Integrate the result of I(t) twice to get I''(t). Replace t by 5 in the result of I"(t) for the final answer. Raghvendra Singh proposed about these exact steps to solve the integral with cos(x) in the integrand instead of sin(x). This approach is worth perhaps monkeying here.
check out
Define the integral as I
convert sin(x)=cos(pi/2-x)
Apply kings property
cos(pi/2-(2pi-x))=cos(-3pi/2+x)
cos(x-3pi/2)
Use the sum/difference
Add two forms of I
Play around using algebra
Split interval of integration into quadrants
Integral on first quadrant leave as it is
In integral on second quadrant use substitution t = \pi-x
In integral on third quadrant use substitution t = x - pi
In integral on fourth quadrant use substitution t = 2\pi - x
Add these integrals to get integral
4\int_{0}^{\pi}{\frac{25+9sin^2(x)}{(25-9sin^2(x))^2}dx}
Use substitution t = tan(x) to get
4\int_{0}^{\infty}{\frac{25+34t^2}{(25+16t^2)^2}}dt}
Now we can rewrite it as
4(\int_{0}^{\infty}{\frac{25+16t^2}{(25+16t^2)^2}}dt} + 18\int_{0}^{\infty}{\frac{t^2}{(25+16t^2)^2}dt})
Then integrate \int_{0}^{\infty}{\frac{t^2}{(25+16t^2)^2}dt} by parts
It is done
Another potential solution:
Consider I(t)=integral 0 to 2π of 1/(t-3sinx).
Calculate I(t) using standard techniques or Weierstrass substitution.
Integrate the result of I(t) twice to get I''(t).
Replace t by 5 in the result of I"(t) for the final answer.
Raghvendra Singh proposed about these exact steps to solve the integral with cos(x) in the integrand instead of sin(x). This approach is worth perhaps monkeying here.
I shall check it out. There are many comments in the que.