Evaluate the following definite integral by letting $u= \pi/2 - x$. The integral is $$\int_0^{\pi/2} \frac{\sin(x)}{\cos(x)+\sin(x)}dx.$$
Asked
Active
Viewed 80 times
0
-
1What did you try so far? – Blazej Jul 21 '15 at 18:51
-
1Welcome to Math SE! What are your thoughts/efforts so far on this problem? You are more likely to get a good response if we know what exactly you are stuck on. – Plutoro Jul 21 '15 at 18:51
-
try the half angle substitution or work with your hint – Dr. Sonnhard Graubner Jul 21 '15 at 18:52
2 Answers
1
Your substitution yields $$I = \int_0^{\pi/2}\frac{\cos u}{\sin u + \cos u} \, \mathrm{d}u = \int_0^{\pi/2}\frac{\cos x}{\sin x + \cos x} \, \mathrm{d}x$$
What can you say about $I + I = 2I$?
Hint: $2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}$.
Zain Patel
- 16,802
-
3
-
-
-
2
-
1@Math-fun, no. The first line is just emphasising that $u$ and $x$ are just dummy variables. – Zain Patel Jul 21 '15 at 19:15
-
1
1
The integral $I$ is given by
$$\begin{align} I&=\int_0^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx \\\\\tag 1 &=\int_0^{\pi/2}\left(1-\frac{\cos x}{\sin x+\cos x}\right)\,dx\\\\ &=\pi/2-\int_0^{\pi/2} \frac{\cos x}{\sin x+\cos x}\,dx \tag 2 \end{align}$$
Enforce the substitution $x\to \pi/2 -x$ in $(1)$. Then, we have
$$I=\int_0^{\pi/2} \frac{\cos x}{\sin x+\cos x}\,dx \tag 3$$
Comparing $(2)$ and $(3)$, we see that
$$\begin{align} I&=\pi/2-I\\\\ &\implies I=\pi/4 \end{align}$$
and we are done!
Mark Viola
- 179,405
-
Please let mw know how I can improve my answer. I really just want to give you the very best answer I can. – Mark Viola Jul 22 '15 at 15:49