Question

How do you solve \[\int{\sqrt{\dfrac{\cos \left( x \right)-{{\cos }^{3}}x}{1-{{\cos }^{3}}x}dx}}\]?

Answer 1

Hint: In the given question, we have been asked to integrate the following function. In order to solve the question, we integrate the numerical by following the substitution method. We substitute \[u={{\cos }^{\dfrac{3}{2}}}\left( x \right)\] and replace it with \[-\dfrac{2}{3}\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}du}\] and solve the numerical by further integration. After we have simplified our sum, we just need to integrate the terms and then we replace the substituted variables by the original variables.

Formula used:
The product to sum formulas of trigonometry:
\[\sin \left( x \right)\sin \left( y \right)=\dfrac{1}{2}\left( \cos \left( y-x \right)-\cos \left( y+x \right) \right)\],
\[{{\sin }^{2}}x=\dfrac{1}{2}\left( 1-\cos \left( 2x \right) \right)\],
\[\cos \left( x \right)\cos \left( y \right)=\dfrac{1}{2}\left( \cos \left( y+x \right)+\cos \left( y-x \right) \right)\],
\[{{\cos }^{2}}x=\dfrac{1}{2}\left( \cos \left( 2x \right)+1 \right)\],
\[\sin \left( x \right)\cos \left( y \right)=\dfrac{1}{2}\left( \sin \left( y+x \right)-\sin \left( y-x \right) \right)\],
\[\cos \left( x \right)\sin \left( y \right)=\dfrac{1}{2}\sin \left( 2x \right)\]
Addition formulas and multiple angle or argument formulas:
\[\sin \left( 3x \right)=3{{\cos }^{2}}\left( x \right)\sin \left( x \right)-{{\sin }^{3}}\left( x \right)\]
\[\cos \left( 3x \right)={{\cos }^{3}}\left( x \right)-3\cos \left( x \right){{\sin }^{2}}\left( x \right)\]

Complete step-by-step solution:
We have the given function,
\[\Rightarrow \int{\sqrt{\dfrac{\cos \left( x \right)-{{\cos }^{3}}x}{1-{{\cos }^{3}}x}dx}}\]
It can be rewrite as,
\[\Rightarrow \int{\dfrac{\sqrt{\cos \left( x \right)-{{\cos }^{3}}x}}{\sqrt{1-{{\cos }^{3}}x}}}dx\]
Applying the product to sum formulas of trigonometry:
\[\sin \left( x \right)\sin \left( y \right)=\dfrac{1}{2}\left( \cos \left( y-x \right)-\cos \left( y+x \right) \right)\],
\[{{\sin }^{2}}x=\dfrac{1}{2}\left( 1-\cos \left( 2x \right) \right)\],
\[\cos \left( x \right)\cos \left( y \right)=\dfrac{1}{2}\left( \cos \left( y+x \right)+\cos \left( y-x \right) \right)\],
\[{{\cos }^{2}}x=\dfrac{1}{2}\left( \cos \left( 2x \right)+1 \right)\],
\[\sin \left( x \right)\cos \left( y \right)=\dfrac{1}{2}\left( \sin \left( y+x \right)-\sin \left( y-x \right) \right)\],
\[\cos \left( x \right)\sin \left( y \right)=\dfrac{1}{2}\sin \left( 2x \right)\]
After applying, we get
\[\Rightarrow \int{\dfrac{\sqrt{\cos \left( x \right)-\dfrac{\cos \left( 3x \right)+3\cos \left( x \right)}{4}}}{\sqrt{1-\dfrac{\cos \left( 3x \right)+3\cos \left( x \right)}{4}}}}dx\]
It can be rewrite as,
\[\Rightarrow \int{\dfrac{\sqrt{\cos \left( x \right)-\cos \left( 3x \right)}}{\sqrt{-\cos \left( 3x \right)-3\cos \left( x \right)+4}}}dx\]
Apply addition formulas and multiple angle or argument formulas:
\[\sin \left( 3x \right)=3{{\cos }^{2}}\left( x \right)\sin \left( x \right)-{{\sin }^{3}}\left( x \right)\]
\[\cos \left( 3x \right)={{\cos }^{3}}\left( x \right)-3\cos \left( x \right){{\sin }^{2}}\left( x \right)\]
After applying, we get
\[\Rightarrow \int{\dfrac{\sqrt{3\cos \left( x \right){{\sin }^{2}}\left( x \right)-{{\cos }^{3}}\left( x \right)+\cos \left( x \right)}}{\sqrt{3\cos \left( x \right){{\sin }^{2}}\left( x \right)-{{\cos }^{3}}\left( x \right)-3\cos \left( x \right)+4}}}dx\]
Rewrite the above function by using trigonometric identities, we get
\[\Rightarrow \int{\dfrac{2\sqrt{\cos \left( x \right)}\sin \left( x \right)}{\sqrt{4-4{{\cos }^{3}}\left( x \right)}}}dx\]
Substitute \[u={{\cos }^{\dfrac{3}{2}}}\left( x \right)\] as,
\[u={{\cos }^{\dfrac{3}{2}}}\left( x \right)\]
Differentiating it, we get
\[\dfrac{du}{dx}=-\dfrac{3\sqrt{\cos \left( x \right)}\sin \left( x \right)}{2}\]
\[dx=-\dfrac{2}{3\sqrt{\cos \left( x \right)}\sin \left( x \right)}du\]
Replacing this in plug in solved integrals,
\[\Rightarrow \int{-\dfrac{2}{3\sqrt{1-{{u}^{2}}}}du}\]
Simplifying the above numerical, we get
\[\Rightarrow -\dfrac{2}{3}\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}du}\]
Now solving,
\[\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}du}\], it is the standard integral which is equal to \[=\arcsin \left( u \right)\]
Replace it in previously solved integral, we get
\[\Rightarrow -\dfrac{2}{3}\int{\dfrac{1}{\sqrt{1-{{u}^{2}}}}du}=-\dfrac{2\arcsin \left( u \right)}{3}\]
Replacing \[u={{\cos }^{\dfrac{3}{2}}}\left( x \right)\], we get
\[\Rightarrow -\dfrac{2\arcsin \left( {{\cos }^{\dfrac{3}{2}}}\left( x \right) \right)}{3}\]
Therefore,
\[\Rightarrow \int{\sqrt{\dfrac{\cos \left( x \right)-{{\cos }^{3}}x}{1-{{\cos }^{3}}x}dx}}=-\dfrac{2\arcsin \left( {{\cos }^{\dfrac{3}{2}}}\left( x \right) \right)}{3}+C\]

Note: Integration is the part of calculus that includes the differentiation. Integration refers to the, add up smaller parts of any area given, volume given, etc to represents the whole value. To solve any given numerical, or function, there are different types of integration methods like integration by substitution, integration by parts, etc.