$2\displaystyle\int_0^\pi \sin x + \sin x \cos x\,dx$

I let $u=\sin x \implies \dfrac{du}{dx}=\cos x \implies dx=\dfrac{du}{\cos x}$

$2\displaystyle\int_{x=0}^{x=\pi} 2u\,du$

When I try to change the limits I just get 0 and 0: Lower limit $=\sin 0 = 0$, upper limit $=\sin \pi=0$

$2\left[u^2\right]^0_0 = 0$. The answer should be 4, not 0.


  • $\begingroup$ What about zeros of $\cos x$ when you deal with $\frac{\mathrm du}{\cos x}$? Why don't you go with the classical $u = \tan\frac x2$? $\endgroup$
    – Ilya
    Apr 4, 2013 at 9:30
  • $\begingroup$ You seem to have forgotten to divide the $\sin x$ term by the $\cos x$ obtained from the substitution. Also, $\sin$ is not bijective on $(0,\pi)$ rendering the application of the substitution theorem invalid. If you want to solve by substitution, it's probably best to split the integration interval into $(0,\pi/2)$ and $(\pi/2,\pi)$. $\endgroup$
    – Lord_Farin
    Apr 4, 2013 at 9:36

3 Answers 3


Assuming the question is : $2\int_{0}^{\pi}(sin(x) + sin(x)cos(x))dx$

It is not 2$\int_{0}^{\pi} 2u. du$ on simplification.

It is 2$\int_{0}^{\pi} udx + udu$

Instead of substituting, you can just divide the integration into two and proceed as follows:

$$=2\int_{0}^{\pi}sinx .dx + \int_{0}^{\pi} sin(2x).dx $$

$$=2(-cos(x)) + \dfrac{(-cos(2x))}{2}$$ Then apply the limit from $0$ to $\pi$:

$$=2(1-(-1)) + 0$$


Hope the answer is clear !

  • $\begingroup$ Now I get it, thank you very much! $\endgroup$ Apr 4, 2013 at 9:48

As $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$

If $f(x)=\sin x+\sin x\cos x,$ $f(\pi+0-x)=\sin(\pi+0-x)+\sin(\pi+0-x)\cos(\pi+0-x)=\sin x-\sin x\cos x$ as $\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$

So, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi (\sin x - \sin x \cos x)dx=I\text{ say}$$

So, $$2I=\int_0^\pi (\sin x + \sin x \cos x)dx+\int_0^\pi (\sin x - \sin x \cos x)dx=2 \int_0^\pi\sin xdx$$

So, $$I=\int_0^\pi\sin xdx=(-\cos x)|_0^\pi=-\cos\pi-(-\cos 0)=2$$

Alternatively, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi \left(\sin x +\frac{\sin2x}2\right)dx=\left(-\cos x -\frac{\cos2x}4\right)_0^\pi$$

$$=\left(\cos x +\frac{\cos2x}4\right)_\pi^0=\cos0+\frac{\cos0}4-\left(\cos\pi+\frac{\cos\pi}4\right)=1+\frac14-\left(-1+\frac14\right)=2$$

In fact, $\int_0^\pi\sin2xdx=\int_0^{2\pi}\sin ydy=(-\cos y)_0^{2\pi}=1-1=0$


You lost the first summand there.

$$ 2\int_0^\pi \sin x + \sin x \cos x\,dx = 2\int_0^\pi \sin x\,dx + 2\int_0^\pi \sin x \cos x\,dx $$ Now integrate both summands separately; you don't need to substitute in the first.

  • $\begingroup$ Even if you are in a class and there is a current subject, you should be ready to use other techniques, or you should realize that the current subject applies only partially. $\endgroup$ Apr 5, 2013 at 1:16

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