Question

Integrate the following function:
${{\sin }^{3}}\left( 2x+1 \right)$

Answer 1

Hint: First of all take (2x + 1) = t and transform the whole integral in terms of t. Now, write ${{\sin }^{3}}t=\left( {{\sin }^{2}}t \right).\left( \sin t \right)$, then write ${{\sin }^{2}}t=1-{{\cos }^{2}}t$ and substitute cos t = y. Now transform the whole integral in terms of y and solve it.

Complete Step-by-step answer:
In this question, we have to integrate the function ${{\sin }^{3}}\left( 2x+1 \right)$. Let us consider the integral given in the question.
$I=\int{{{\sin }^{3}}\left( 2x+1 \right)}\text{ }dx....\left( i \right)$
Let us assume 2x + 1 = t….(ii)
We know that $\dfrac{d}{da}\left( a \right)=1$
So, by differentiating both the sides of the above equation, we get,
2dx = dt
By substituting (2x + 1) = t and $dx=\dfrac{dt}{2}$ in equation (i), we get,
$I=\int{\left( {{\sin }^{3}}t \right)\dfrac{dt}{2}}$
We know that ${{a}^{m+n}}={{a}^{m}}.{{a}^{n}}$. So by using this in the above equation, we can write the above equation as,
$I=\dfrac{1}{2}\int{\left( {{\sin }^{2}}t \right)}\left( \sin t \right)dt$
We know that ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$. By using this in the above equation, we get,
$I=\dfrac{1}{2}\int{\left( 1-{{\cos }^{2}}t \right)\left( \sin t \right)dt....\left( iii \right)}$
Now let us assume cos t = y… (iv)
Now, we know that $\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta$
So by differentiating the above equation, we get
$-\sin t\text{ }dt=dy$
$\sin t\text{ }dt=-dy$
So, by substituting cos t = y and sin t dt = – dy in equation (iii), we get,
$I=\dfrac{1}{2}\int{\left( 1-{{y}^{2}} \right)\left( -dy \right)}$
$I=\dfrac{1}{2}\int{\left( {{y}^{2}}-1 \right)dy}$
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$. So, by using this in the above equation, we get,
$I=\dfrac{1}{2}\left( \dfrac{{{y}^{3}}}{3}-y \right)+C$
Now, by substituting the value of y = cos t from equation (iv) in the above equation, we get,
$I=\dfrac{1}{2}\left[ \dfrac{{{\left( \cos t \right)}^{3}}}{3}-\cos t \right]+C$
Now by again substituting t = (2x + 1) from equation (ii) in the above equation, we get,
$I=\dfrac{1}{2}\left[ \dfrac{{{\left( \cos \left( 2x+1 \right) \right)}^{3}}}{3}-\cos \left( 2x+1 \right) \right]+C$

Note: In these types of questions, students must remember to convert the new variable to the original variable at the end of the solution, step by step beginning from the last one. Also, students can cross-check their answer by differentiating it and checking if it is coming equal to the given question or not because we know that differentiation and integration are the opposite of each other.