Is it possible to compute $\int_0^x\frac{t^2dt}{(t\sin(t)+\cos(t))^2}$?

Question

One of my students gave this integrale. By parts makes it more complicate. I see no classical substitution. Here is the integral to find $$\int_0^x\frac{t^2dt}{(t\sin(t)+\cos(t))^2}$$

Thanks in advance.

Answer 1

The required integral is: $$I=\int_0^x\frac{t^2dt}{(t\sin(t)+\cos(t))^2}$$

First we rewrite it as: $$I=\int_0^x\frac{t\sin(t)dt}{t\sin(t)+\cos(t)}(I_1)+\int_0^x\frac{t\cos(t)(t\cos(t)-\sin(t))dt}{(t\sin(t)+\cos(t))^2}(I_2)$$

Now apply integration by parts in $I_2$ by taking $$f(t)=t\sin(t)-\cos(t)$$ as the first function and $$g(t)=\frac{t\cos(t)}{(t\sin(t)+\cos(t))^2}$$ as the second function. You will then get $$I_1=\int_0^x\frac{\sin(t)-t\cos(t)dt}{t\sin(t)+\cos(t)}-I_2$$

This implies that $$I=I_1+I_2=\frac{\sin(x)-x\cos(x)}{x\sin(x)+\cos(x)}-I_2+I_2=\frac{\sin(x)-x\cos(x)}{x\sin(x)+\cos(x)}$$

Answer 2

Use the antiderivative: $$ \int\frac{t^2dt}{(t \sin (t)+\cos (t))^2}=\frac{\sin (t)-t \cos (t)}{t \sin (t)+\cos (t)}+C. $$