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Question: How do you integrate \( \int \dfrac{{\sin x}}{{{{(2 + 3\cos x)}^2}}}dx \) using substitution?...

How do you integrate ∫sin⁡x(2+3cos⁡x)2dx\int \dfrac{{\sin x}}{{{{(2 + 3\cos x)}^2}}}dx using substitution?

Explanation

Solution

Hint : To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value. Here we put the denominator part as u and solve further.

Complete step by step solution:
The given expression: ∫sin⁡x(2+3cos⁡x)2dx\int \dfrac{{\sin x}}{{{{(2 + 3\cos x)}^2}}}dx
Let u=2+3cos⁡xu = 2 + 3\cos x
Differentiate the above equation that we supposed:
⇒du=−3sin⁡xdx\Rightarrow du = - 3\sin xdx
⇒sin⁡xdx=−13du\Rightarrow \sin xdx = - \dfrac{1}{3}du
Now, put the upper values in the main given expression:
∫sin⁡x(2+3cos⁡x)2dx =−13∫duu2 =−13∫u−2du =13.1u =13.(12+3cos⁡x)+C   \int \dfrac{{\sin x}}{{{{(2 + 3\cos x)}^2}}}dx \\\ = - \dfrac{1}{3}\int \dfrac{{du}}{{{u^2}}} \\\ = - \dfrac{1}{3}\int {u^{ - 2}}du \\\ = \dfrac{1}{3}.\dfrac{1}{u} \\\ = \dfrac{1}{3}.(\dfrac{1}{{2 + 3\cos x}}) + C \;
So, the correct answer is “13.(12+3cos⁡x)+C \dfrac{1}{3}.(\dfrac{1}{{2 + 3\cos x}}) + C”.

Note : Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.