Question

Solve the following integration

$\int \: \frac{ {sin}^{2} x}{ {4cos}^{2} x + {sin}^{2} x} dx$
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm \: \displaystyle\int\rm \frac{ {sin}^{2} x}{ {4cos}^{2} x + {sin}^{2} x} \: dx$

can be further rewritten as

$\rm \: \displaystyle\int\rm \frac{ {sin}^{2} x}{ {4(1 - sin}^{2} x) + {sin}^{2} x} \: dx$

$\rm \: \displaystyle\int\rm \frac{ {sin}^{2} x}{ {4 - 4sin}^{2} x + {sin}^{2} x} \: dx$

$\rm \: \displaystyle\int\rm \frac{ {sin}^{2} x}{ {4 - 3sin}^{2} x} \: dx$

$\rm \: - \dfrac{1}{3} \displaystyle\int\rm \frac{ - 3{sin}^{2} x}{ {4 - 3sin}^{2} x} \: dx$

$\rm \: - \dfrac{1}{3} \displaystyle\int\rm \frac{4 - 3{sin}^{2} x - 4}{ {4 - 3sin}^{2} x} \: dx$

$\rm \: - \dfrac{1}{3} \displaystyle\int\rm \bigg(1 - \frac{4}{ {4 - 3sin}^{2} x}\bigg) \: dx$

$\rm \: = \: - \dfrac{1}{3} \displaystyle\int\rm 1 \: dx \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{1}{ {4 - 3sin}^{2} x} \: dx$

In second integral, divide numerator and denominator by cos^2x

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{ {sec}^{2}x }{ {4 {sec}^{2}x - 3tan}^{2} x} \: dx$

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{ {sec}^{2}x }{ {4 {(1 + tan}^{2}x) - 3tan}^{2} x} \: dx$

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{ {sec}^{2}x }{ { {4 + 4tan}^{2}x - 3tan}^{2} x} \: dx$

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{ {sec}^{2}x }{4 + {tan}^{2} x} \: dx$

Now, to solve this integral, we use method of Substitution.

So, Substitute

$\rm \: tanx = y$

$\rm \: {sec}^{2} x \: dx \: = \: dy$

So, on substituting the values, we get

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{ dy }{4 + {y}^{2} } \:$

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \displaystyle\int\rm \frac{ dy }{{y}^{2} + {2}^{2} } \:$

$\rm \: = \: - \dfrac{1}{3} x \: + \: \dfrac{4}{3} \times \dfrac{1}{2} {tan}^{ - 1} \dfrac{y}{2} + c \:$

$\rm \: = \: - \: \dfrac{x}{3}\: + \: \dfrac{2}{3} {tan}^{ - 1} \bigg(\dfrac{tanx}{2}\bigg) + c \:$

Hence,

$\rm \: \displaystyle\int\rm \frac{ {sin}^{2} x}{ {4cos}^{2}x + {sin}^{2}x}dx = \: - \: \dfrac{x}{3}\: + \: \dfrac{2}{3} {tan}^{ - 1} \bigg(\dfrac{tanx}{2}\bigg) + c \:$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$