Evaluate $ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $ using substitution $t=\frac{\pi}{2}-x$

Question

This problem is given on a sample test for my calculus two class. $$ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $$ I can find the value of this integral using other substitutions which lead to partial fractions but the prof added a hint to use the substitution $$t=\frac{\pi}{2}-x $$ so I've been trying to figure out how to do it his way but am pretty lost. Any ideas?

Answer 1

Hint: Do the substitution, and stop. You will get that our integral is equal to another integral over the same interval. Add them!

Symmetry solves problems.

Answer 2

HINT: $\sin\left(\frac{\pi}2-x\right)=\cos x$, and $\cos\left(\frac{\pi}2-x\right)=\sin x$. Look at the sum of your original integral and the new one.

Answer 3

$I = \int_0^{\pi/2} \frac{\sin x}{\sin x+ \cos x}\mathrm{d}x$. On the other hand, with the substitution $t = \frac{\pi}{2} - x$, we obtain \begin{align} I & = \int_0^{\pi/2} \frac{\sin (\frac{\pi}{2}-t)}{\sin (\frac{\pi}{2}-t)+ \cos (\frac{\pi}{2}-t)}\mathrm{d}t \\ & = \int_0^{\pi/2} \frac{\cos t}{\sin t+ \cos t}\mathrm{d}t \\ & = \int_0^{\pi/2} \frac{\cos x}{\sin x+ \cos x}\mathrm{d}x \\ \end{align} as $\sin(\frac{\pi}{2}-t) = \cos t$ and $\cos(\frac{\pi}{2}-t) = \sin t$.

Adding the "original" form of $I$ and this new form, we have \begin{align} 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x+ \cos x}\mathrm{d}x = \int_0^{\pi/2} \mathrm{d}x = \frac{\pi}{2}, \end{align} from which we obtain $I = \frac{\pi}{4}$, and say to ourselves, "Hmm that is cool."

Answer 4

There is an easy way to solve. Let $$ A=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}dx, B=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx. $$ Then $$ A+B=\frac{\pi}{2} $$ and $$ A-B=\int_0^{\frac{\pi}{2}}\frac{\sin x-\cos x}{\sin x+\cos x}dx=\int_0^{\frac{\pi}{2}}\frac{-d(\sin x+\cos x)}{\sin x+\cos x}=-\ln(\sin x+\cos x)\big|_0^{\frac{\pi}{2}}=0. $$ From this, we have $$ A=B=\frac{\pi}{4}. $$ Generally, I can use this method to compute (for $a,b>0$) $$ A=\int_0^{\frac{\pi}{2}}\frac{\sin x}{a\sin x+b\cos x}dx, B=\int_0^{\frac{\pi}{2}}\frac{\cos x}{a\sin x+b\cos x}dx. $$ It is easy to see $$ aA+bB=\frac{\pi}{2} $$ and \begin{eqnarray*} aB-bA&=&\int_0^{\frac{\pi}{2}}\frac{-b\sin x+a\cos x}{a\sin x+b\cos x}dx\\ &=&\int_0^{\frac{\pi}{2}}\frac{d(a\sin x+b\cos x)}{a\sin x+a\cos x}\\ &=&\ln(a\sin x+b\cos x)\big|_0^{\frac{\pi}{2}}\\ &=&\ln a-\ln b. \end{eqnarray*} From this, we have $$ A=\frac{a\pi-2b\ln\frac{a}{b}}{2(a^2+b^2)},B=\frac{b\pi+2a\ln\frac{a}{b}}{2(a^2+b^2)}. $$

Answer 5

Rationalization $$ \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} \cdot \frac{\cos x-\sin x}{\cos x-\sin x} d x \\ & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sin x \cos x-\sin ^2 x}{\cos 2 x} d x\\&= \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sin 2 x-(1-\cos 2 x)}{\cos 2 x} d x\\ & =\frac{1}{2} \left[-\frac{1}{2} \ln (\cos 2 x)-\frac{1}{2} \ln (\sec 2 x+\tan 2 x)+x\right]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{4} \end{aligned} $$