0
$\begingroup$

$$\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}$$

I have tried it using transformation but its getting lengthy..it must be short question. please help

$\endgroup$
1
$\begingroup$

Hint use $f (x)=f (a+b-x ) $ where a,b are lower ,upper limits . Then everything is straightforward

$\endgroup$
0
$\begingroup$

Using the hint given by Archis Welankar: $$\int_a^b f(x) dx=\int_a^bf(a+b-x)dx;\\ \int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx=\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin (\pi/2-x)}} {\sqrt[3]{\sin (\pi/2-x)} + \sqrt[3]{\cos (\pi/2-x)}}dx \Rightarrow \\ \underbrace{\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx}_{A}=\underbrace{\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\cos x}} {\sqrt[3]{\cos x} + \sqrt[3]{\sin x}}dx}_{B} \Rightarrow \\ A-B=0.$$ Also: $$\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx=\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}+\sqrt[3]{\cos x}-\sqrt[3]{\cos x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx=\\ \int_{\pi/6}^{\pi/3} 1-\frac {\sqrt[3]{\cos x}} {\sqrt[3]{\cos x} + \sqrt[3]{\sin x}}dx \Rightarrow \\ A+B=\frac{\pi}{6}.$$ Hence: $$A=\frac{\pi}{12}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.