Question

The value of the definite integral $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx$ is?

A. $\dfrac{\pi -2}{4}$

B. $\dfrac{\pi -2}{8}$

C. $\dfrac{\pi -1}{4}$

D. $\dfrac{\pi -1}{2}$

Answer 1

Hint: To solve this integral, we should apply a property of the definite integrals. The property is $\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)}}dx$. Using this property in the integral, we get $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}\left( \dfrac{\pi }{2}-x \right)}{\sin \left( \dfrac{\pi }{2}-x \right)+\cos \left( \dfrac{\pi }{2}-x \right)}}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{3}}x}{\sin x+\cos x}}dx$. By adding the two integrals $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx$ and $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{3}}x}{\sin x+\cos x}}dx$ , we get $2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{\sin x+\cos x}}dx$. By using the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$, we can solve the integral.

Complete step-by-step answer:

We are given the integral $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx$.

We know a formula in integration $\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)}}dx$.

Here in the question $a=0,b=\dfrac{\pi }{2}$, $f\left( x \right)=\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}$

By applying the formula, we get

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}\left( \dfrac{\pi }{2}-x \right)}{\sin \left( \dfrac{\pi }{2}-x \right)+\cos \left( \dfrac{\pi }{2}-x \right)}}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{3}}x}{\sin x+\cos x}}dx$

We got the two expressions for the integral I as

$ I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx\to \left( 1 \right) $

$ I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{3}}x}{\sin x+\cos x}}dx\to \left( 2 \right) $

By adding equations 1 and 2, we get

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{3}}x}{\sin x+\cos x}}dx$

As the limits of the two definite integrals are same, we can write the two integrals as a single integral

$\int\limits_{a}^{b}{f\left( x \right)dx+}\int\limits_{a}^{b}{g\left( x \right)dx=\int\limits_{a}^{b}{\left( f\left( x \right)+g\left( x \right) \right)}}dx$

Using this property, we get

$ 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\sin }^{3}}x}{\sin x+\cos x}+\dfrac{{{\cos }^{3}}x}{\sin x+\cos x} \right)}dx $

$ 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{\sin x+\cos x}dx} $

We know the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

By taking $a=\sin x,b=\cos x$, we can write the numerator in the integral as

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( \sin x+\cos x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x-\sin x\cos x \right)}{\sin x+\cos x}dx}$

By cancelling $\sin x+\cos x$, we get

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x+{{\cos }^{2}}x-\sin x\cos x \right)dx}$

We know the formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1-\sin x\times \cos x \right)dx}$

We know that

$ \sin 2x=2\sin x\times \cos x $

$  \dfrac{\sin 2x}{2}=\sin x\times \cos x $

The integral can be written as

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1-\dfrac{\sin 2x}{2} \right)dx}$

We know the integration formula $\int{\sin nxdx=\dfrac{-\cos nx}{n}}$

By integrating the function inside the integral, we get

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1-\dfrac{\sin 2x}{2} \right)dx}=\left[ x-\left( -\dfrac{\cos 2x}{2\times 2} \right) \right]_{0}^{\dfrac{\pi }{2}}=\left[ x+\dfrac{\cos 2x}{4} \right]_{0}^{\dfrac{\pi }{2}}$

By applying the limits, we get

$2I=\left( \dfrac{\pi }{2}+\dfrac{\cos 2\times \dfrac{\pi }{2}}{4} \right)-\left( 0+\dfrac{\cos 2\times 0}{4} \right)=\dfrac{\pi }{2}+\dfrac{-1}{4}-\dfrac{1}{4}=\dfrac{\pi }{2}-\dfrac{1}{2}=\dfrac{\pi -1}{2}$

Dividing by 2, we get

$I=\dfrac{\pi -1}{4}$

$\therefore $ The value of the integral $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{3}}x}{\sin x+\cos x}}dx=\dfrac{\pi -1}{4}$. 

Hence, the correct answer is option (C).

Note: Students might commit mistakes by forgetting the value of 2 in $2I$ and get an answer as $I=\dfrac{\pi -1}{2}$. There is option-D which contains the wrong option. Students, without dividing by 2, opt for the option-D. To avoid this mistake, it is always a good practice to write both L.H.S and R.H.S in every step even while doing the rough work. This method of writing eliminates the error.