Question

Evaluate the given function and fill in the blanks: $\int{\sqrt{\dfrac{\cos x-{{\cos }^{3}}x}{1-{{\cos }^{3}}x}}dx=\_\_\_\_\_\_\_\_\_+C}$
$\left( \text{A} \right)\text{ }\dfrac{2}{3}{{\cos }^{-1}}\left( {{\sin }^{\dfrac{3}{2}}}x \right)$
$\left( B \right)\text{ }\dfrac{2}{3}{{\tan }^{-1}}\left( {{\cos }^{\dfrac{3}{2}}}x \right)$
$\left( \text{C} \right)-\dfrac{2}{3}{{\sin }^{-1}}\left( {{\cos }^{\dfrac{3}{2}}}x \right)$
$\left( D \right)\text{ }\dfrac{2}{3}{{\sin }^{-1}}\left( {{\sin }^{\dfrac{3}{2}}}x \right)$

Answer 1

Hint: In this question we have been given with a question of indefinite integrals which we have to solve. We will first take $\cos x$ common in the numerator and convert it into the form of $\sin x$. We will then use appropriate substitution to get the expression in a simplified manner. After using the formula for integrating we will get the required solution and we will check which option is the correct solution.

Complete step-by-step solution:
We have the integral given to us as:
\[\Rightarrow \int{\sqrt{\dfrac{\cos x-{{\cos }^{3}}x}{1-{{\cos }^{3}}x}}}dx\]
In the numerator, we can see that the term $\cos x$ is common therefore on taking it out, we get:
\[\Rightarrow \int{\sqrt{\dfrac{\cos x\left( 1-{{\cos }^{2}}x \right)}{1-{{\cos }^{3}}x}}}dx\]
Now we know the trigonometric formula that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ which implies $1-{{\cos }^{2}}x={{\sin }^{2}}x$ therefore, on substituting it in the expression, we get:
\[\Rightarrow \int{\sqrt{\dfrac{\cos x\left( {{\sin }^{2}}x \right)}{1-{{\cos }^{3}}x}}}dx\]
Now in the numerator, we have a term in square under the square root therefore on taking it out, we get:
\[\Rightarrow \int{\sqrt{\dfrac{\cos x}{1-{{\cos }^{3}}x}}\times \sin xdx}\]
Now we know that square root can be written in the exponent form as ${{a}^{\dfrac{1}{2}}}$ therefore the numerator can be written as:
\[\Rightarrow \int{\dfrac{{{\cos }^{\dfrac{1}{2}}}x\cdot \sin x}{\sqrt{1-{{\cos }^{3}}x}}dx}\]
Now the term ${{\cos }^{3}}x$ in the denominator can be written as \[{{\left( {{\cos }^{\dfrac{3}{2}}}x \right)}^{2}}\] because according to the law of exponents nothing it being changed.
On substituting, we get:
\[\Rightarrow \int{\dfrac{{{\cos }^{\dfrac{1}{2}}}x\cdot \sin x}{\sqrt{1-{{\left( {{\cos }^{\dfrac{3}{2}}}x \right)}^{2}}}}dx}\]
Now on substituting $t={{\cos }^{\dfrac{3}{2}}}x$
On differentiating, we get $dt=\dfrac{3}{2}{{\cos }^{\dfrac{1}{2}}}x\left( -\sin x \right)dx$
On rearranging the terms, we get $-\dfrac{2}{3}dt={{\cos }^{\dfrac{1}{2}}}x\sin xdx$.
On substituting it in the expression, we get:
\[\Rightarrow -\dfrac{2}{3}\int{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}\]
Now we know that $\int{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)}+C$ therefore on integrating, we get:
\[\Rightarrow -\dfrac{2}{3}{{\sin }^{-1}}\left( \dfrac{t}{1} \right)+C\]
On resubstituting the value of $t$, we get:
\[\Rightarrow -\dfrac{2}{3}{{\sin }^{-1}}\left( {{\cos }^{\dfrac{3}{2}}}x \right)+C\], which is the required solution therefore the correct option is $\left( C \right)$.

Note: In this question we used the substitution method to solve the integral. It is to be remembered that in an indefinite integral there are no limiting conditions. It is also to be remembered that integration and derivatives are opposite of each other. If the integration of a term $a$ is $b$, then the derivative of the term $b$ will be $a$.