Question

Evaluate :-
$\\ \color{brown} \pmb{\boxed{ \tt\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\sin x \cos x} d x}}$
$\\ \\ \rule{300pt}{1pt}$
$\: \:$


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Answer 1

$\rm \: \: \: \: \: \dag \: \huge{\underline{\underline{Given:}}}$

$\sf \huge \red{\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\sin x \cos x} \: dx}$

$\huge\colorbox{white}{\colorbox{white}{\colorbox{white}{ Solution }}}$

$\begin{gathered}\sf Let \: I = \large {\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\sin x \cos x} \: dx} \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow(1) \end{gathered}$

$\begin{gathered}\sf \therefore \: \: I = \large {\int_{0}^{\frac{\pi}{2}} \dfrac{\sin ^{2} \bigg( \large\frac{ \pi}{2} - x\bigg) }{1+\sin \bigg ( \frac{\pi}{2 \: } - x \bigg) \cos \bigg( \dfrac{\pi}{2} - x \bigg)} \: dx} \end{gathered}$

$\begin{gathered}\sf \qquad\bigg[ {\int_{0}^a \sf f(x)\: dx \: = \int_{0}^a } f (a \: - \: x)dx\bigg]\end{gathered}$

$\begin{gathered}\sf \therefore \: \: I = \large {\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x \: dx}{1+\cos x \: \sin x} \: dx} \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow(2) \end{gathered}$

On adding eqn (1) and eqn (2) , we have

$\begin{gathered}\sf\qquad \: 2 I = \large {\int_{0}^{\frac{\pi}{2}} \frac{(\sin ^{2} x \: + \: cos^{2}x )dx}{1+\sin x \cos x \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

$\begin{gathered}\sf \qquad = \large {\int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\sin x \cos x} \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow(1) \end{gathered}$

$\rm \: Divided \: Numerator \: and \: denominator \: by \: cos ^2 x \: ;We \: have$

$\begin{gathered}\sf\qquad \: 2 I = \large {\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x \: dx}{\sec^{2}x \: + \tan x \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

$\rm \large \: put \: tan \: x \: = \: t \rightarrow {sec}^{2} \: x \: \: dx = dx$

$\rm \large \: when \: x \: = \: 0 \: \rightarrow \: t \: = \: 0$

$\begin{gathered} \sf when \: \: x = \: \frac{ \pi}{2} \: \longrightarrow \: t \: = \infty \end{gathered}$

$\begin{gathered}\sf \qquad \therefore \: 2 I = \large {\int_{0}^{ \infty} \frac{dt}{ {t}^{2} \: + \: t \: + 1 \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

$\begin{gathered}\sf\qquad \implies \large {\int_{0}^{ \infty} \frac{dt}{ {t}^{2} \: + \: t \: + \frac{1}{4} \: + \: \frac{3}{4} \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

$\begin{gathered}\sf\qquad \implies \large {\int_{0}^{ \infty} \frac{dt}{ \bigg (t \: + \: \frac{1}{2} \bigg)^{2} \: + \bigg( \frac{ \sqrt3}{2} \bigg)^{2} \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

$\begin{gathered}\sf\qquad \implies \large \frac{2}{ \sqrt3} \: \bigg[ \bigg( \tan \: \frac{t \: + \: 1}{ \sqrt3} \bigg) \bigg ]_ 0^{\infty} \end{gathered}$

$\begin{gathered}\sf\qquad \implies \large \frac{1}{ \sqrt3} \: \: \bigg[ \frac{ \pi}{2} \: - \: \frac{ \pi}{ 6} \bigg ] \end{gathered}$

$\begin{gathered}\sf\qquad \implies \large \frac{1}{ \sqrt3} \: \: \bigg( \frac{3 \pi \:- \: \pi}{6} \: \: \bigg ) \end{gathered}$

$\begin{gathered}\sf\qquad \implies \large \purple{\frac{ \pi}{ 3\sqrt3}} \: \: \end{gathered}$

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Answer 2

Answer:

\rm \: \: \: \: \: \dag \: \huge{\underline{\underline{Given:}}}†

Given:

\sf \huge \red{\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\sin x \cos x} \: dx}∫

0

2

π

1+sinxcosx

sin

2

x

dx

\huge\colorbox{white}{\colorbox{white}{\colorbox{white}{ Solution }}}

Solution

\begin{gathered} \begin{gathered}\sf Let \: I = \large {\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\sin x \cos x} \: dx} \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow(1) \end{gathered} \\ \\ \end{gathered}

LetI=∫

0

2

π

1+sinxcosx

sin

2

x

dx⟶(1)

\begin{gathered} \begin{gathered}\sf \therefore \: \: I = \large {\int_{0}^{\frac{\pi}{2}} \dfrac{\sin ^{2} \bigg( \large\frac{ \pi}{2} - x\bigg) }{1+\sin \bigg ( \frac{\pi}{2 \: } - x \bigg) \cos \bigg( \dfrac{\pi}{2} - x \bigg)} \: dx} \end{gathered} \\ \end{gathered}

∴I=∫

0

2

π

1+sin(

2

π

−x)cos(

2

π

−x)

sin

2

(

2

π

−x)

dx

\begin{gathered}\sf \qquad\bigg[ {\int_{0}^a \sf f(x)\: dx \: = \int_{0}^a } f (a \: - \: x)dx\bigg]\end{gathered}

[∫

0

a

f(x)dx=∫

0

a

f(a−x)dx]

\begin{gathered} \begin{gathered}\sf \therefore \: \: I = \large {\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x \: dx}{1+\cos x \: \sin x} \: dx} \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow(2) \end{gathered} \\ \\ \end{gathered}

∴I=∫

0

2

π

1+cosxsinx

cos

2

xdx

dx⟶(2)

On adding eqn (1) and eqn (2) , we have

\begin{gathered} \begin{gathered}\sf\qquad \: 2 I = \large {\int_{0}^{\frac{\pi}{2}} \frac{(\sin ^{2} x \: + \: cos^{2}x )dx}{1+\sin x \cos x \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} \\ \\ \end{gathered}

2I=∫

0

2

π

1+sinxcosx

(sin

2

x+cos

2

x)dx

\begin{gathered}\begin{gathered}\sf \qquad = \large {\int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\sin x \cos x} \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow(1) \end{gathered} \\ \\ \end{gathered}

=∫

0

2

π

1+sinxcosx

dx

⟶(1)

\rm \: Divided \: Numerator \: and \: denominator \: by \: cos ^2 x \: ;We \: haveDividedNumeratoranddenominatorbycos

2

x;Wehave

\begin{gathered} \begin{gathered}\sf\qquad \: 2 I = \large {\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x \: dx}{\sec^{2}x \: + \tan x \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} \\ \\ \end{gathered}

2I=∫

0

2

π

sec

2

x+tanx

sec

2

xdx

\rm \large \: put \: tan \: x \: = \: t \rightarrow {sec}^{2} \: x \: \: dx = dxputtanx=t→sec

2

xdx=dx

\rm \large \: when \: x \: = \: 0 \: \rightarrow \: t \: = \: 0whenx=0→t=0

\begin{gathered} \sf when \: \: x = \: \frac{ \pi}{2} \: \longrightarrow \: t \: = \infty \end{gathered}

whenx=

2

π

⟶t=∞

\begin{gathered} \begin{gathered}\sf \qquad \therefore \: 2 I = \large {\int_{0}^{ \infty} \frac{dt}{ {t}^{2} \: + \: t \: + 1 \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} \\ \\ \end{gathered}

∴2I=∫

0

∞

t

2

+t+1

dt

\begin{gathered} \begin{gathered}\sf\qquad \implies \large {\int_{0}^{ \infty} \frac{dt}{ {t}^{2} \: + \: t \: + \frac{1}{4} \: + \: \frac{3}{4} \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} \\ \\ \end{gathered}

⟹∫

0

∞

t

2

+t+

4

1

+

4

3

dt

\begin{gathered}\begin{gathered}\sf\qquad \implies \large {\int_{0}^{ \infty} \frac{dt}{ \bigg (t \: + \: \frac{1}{2} \bigg)^{2} \: + \bigg( \frac{ \sqrt3}{2} \bigg)^{2} \:} \:} \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered} \\ \\ \end{gathered}

⟹∫

0

∞

(t+

2

1

)

2

+(

2

3

)

2

dt

\begin{gathered}\begin{gathered}\sf\qquad \implies \large \frac{2}{ \sqrt3} \: \bigg[ \bigg( \tan \: \frac{t \: + \: 1}{ \sqrt3} \bigg) \bigg ]_ 0^{\infty} \end{gathered} \\ \\ \end{gathered}

⟹

3

2

[(tan

3

t+1

)]

0

∞

\begin{gathered}\begin{gathered}\sf\qquad \implies \large \frac{1}{ \sqrt3} \: \: \bigg[ \frac{ \pi}{2} \: - \: \frac{ \pi}{ 6} \bigg ] \end{gathered} \\ \\ \end{gathered}

⟹

3

1

[

2

π

−

6

π

]

\begin{gathered}\begin{gathered}\sf\qquad \implies \large \frac{1}{ \sqrt3} \: \: \bigg( \frac{3 \pi \:- \: \pi}{6} \: \: \bigg ) \end{gathered} \\ \\ \end{gathered}

⟹

3

1

(

6

3π−π

)

\begin{gathered}\begin{gathered}\sf\qquad \implies \large \purple{\frac{ \pi}{ 3\sqrt3}} \: \: \end{gathered} \\ \\ \end{gathered}

⟹

3

3

π

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