Question

Integrate the following
$\displaystyle \sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2sin2x} } = (tan \: x) {}^{a} + c(tan \: x) {}^{b} + k$
Here,
k is the constant of integration

Find the value of a + b + c.

Thank You​

Answer 1

GIVEN :–

$\\ \implies \displaystyle \sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2sin2x} } = (tan \: x) {}^{a} + c(tan \: x) {}^{b} + k$

TO FIND :–

• a + b + c = ?

SOLUTION :–

$\\ I = \displaystyle\sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2sin2x} }$

• Using identity –

$\\ \longrightarrow \sf \: \sin(2x) = 2 \sin(x) \cos(x)$

• So that –

$\\ I = \displaystyle\sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2(2 \sin x \cos x)} }$

$\\ I = \displaystyle\sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{4 \sin x \cos x} }$

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{ {cos}^{3}x \sqrt{ \sin(x) } \sqrt{ \cos(x) } }$

• We should write this as –

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{( {cos}^{3}x ) \sqrt{ \sin(x) } \sqrt{ \cos(x) } \left( \dfrac{ \sqrt{ \cos(x) } }{ \sqrt{ \cos(x) } } \right) }$

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{( {cos}^{3}x ) \sqrt{ \cos(x) } \sqrt{ \cos(x) } \left( \dfrac{ \sqrt{ \sin(x) } }{ \sqrt{ \cos(x) } } \right) }$

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{( {cos}^{4}x ) \sqrt{ \tan(x) }}$

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{( { \sec}^{4}x ) dx}{ \sqrt{ \tan(x) }}$

• We should write this as –

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{( { \sec}^{2}x ).( { \sec}^{2}x ) dx}{ \sqrt{ \tan(x) }}$

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{(1 + { \tan}^{2}x ).( { \sec}^{2}x ) dx}{ \sqrt{ \tan(x) }} \: \: \: \: \left[ \: \because \: \: { \sec}^{2}x =1 + { \tan}^{2}x \right]$

• Let put tan(x) = t –

• Differentiate with respect to 't' –

$\\ \implies \sf { \sec}^{2}(x) \dfrac{dx}{dt} = \dfrac{dt}{dt}$

$\\ \implies \sf { \sec}^{2}(x) \dfrac{dx}{dt} = 1$

$\\ \implies \sf { \sec}^{2}(x) dx = dt$

• So that –

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{(1 + {t}^{2} ).dt}{ \sqrt{t }}$

$\\ I = \displaystyle\sf \dfrac{1}{2} \int \left( \dfrac{1}{ \sqrt{t }} + {t}^{ \frac{3}{2}} \right) .dt$

• Using formula –

$\\ \longrightarrow \displaystyle\sf \int {x}^{n} .dx = \dfrac{ {x}^{n + 1}}{n + 1}$

• So –

$\\ I = \sf \dfrac{1}{2} \left( \dfrac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + \dfrac{{t}^{ \frac{5}{2}}}{ \frac{5}{2} } \right) + k$

$\\ I = \sf \dfrac{1}{2} \left( 2 \times { {t}^{ \frac{1}{2} } } + {t}^{ \frac{5}{2}} \times { \frac{2}{5} } \right) + k$

$\\ I = \sf {t}^{ \frac{1}{2} } + \frac{1}{5} {t}^{ \frac{5}{2} } + k$

• Now compare –

$\\ \: \: \: \: { \huge{.}} \sf \: \: a = \dfrac{1}{2}$

$\\ \: \: \: \: { \huge{.}} \sf \: \: b = \dfrac{5}{2}$

$\\ \: \: \: \: { \huge{.}} \sf \: \: c = \dfrac{1}{5}$

$\\ \implies \sf a + b + c = \dfrac{1}{2} + \dfrac{5}{2} +\dfrac{1}{5}$

$\\ \implies \sf a + b + c =3 + \dfrac{1}{5}$

$\\ \implies \sf a + b + c = \dfrac{15 + 1}{5}$

$\\ \implies \large{ \boxed{ \sf a + b + c = \dfrac{16}{5}}}$

Answer 2

$\displaystyle \sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2sin2x} } = (tan \: x) {}^{a} + c(tan \: x) {}^{b} + k$

Find a+b+c=?

=>$I = \displaystyle\sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2sin2x} }$

By using identity:-

→ $\sf \: \sin(2x) = 2 \sin(x) \cos(x)$

Therefore, $I = \displaystyle\sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{2(2 \sin x \cos x)} }$

$I = \displaystyle\sf \int \dfrac{dx}{ {cos}^{3}x \sqrt{4 \sin x \cos x} }$

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{ {cos}^{3}x \sqrt{ \sin(x) } \sqrt{ \cos(x) } }$

We can write it as:-

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{( {cos}^{3}x ) \sqrt{ \sin(x) } \sqrt{ \cos(x) } \left( \dfrac{ \sqrt{ \cos(x) } }{ \sqrt{ \cos(x) } } \right) }$

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{( {cos}^{3}x ) \sqrt{ \cos(x) } \sqrt{ \cos(x) } \left( \dfrac{ \sqrt{ \sin(x) } }{ \sqrt{ \cos(x) } } \right) }$

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{dx}{( {cos}^{4}x ) \sqrt{ \tan(x) }}$

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{( { \sec}^{4}x ) dx}{ \sqrt{ \tan(x) }}$

We can write it as:-

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{( { \sec}^{2}x ).( { \sec}^{2}x ) dx}{ \sqrt{ \tan(x) }}$

As, ${ \sec}^{2}x =1 + { \tan}^{2}x$

$I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{(1 + { \tan}^{2}x ).( { \sec}^{2}x ) dx}{ \sqrt{ \tan(x) }}$

Let tan(x) = t

=> $\sf { \sec}^{2}(x) \dfrac{dx}{dt} = \dfrac{dt}{dt}$

=> $\sf { \sec}^{2}(x) \dfrac{dx}{dt} = 1$

=> $\sf { \sec}^{2}(x) dx = dt$

Therefore, $I = \displaystyle\sf \dfrac{1}{2} \int \dfrac{(1 + {t}^{2} ).dt}{ \sqrt{t }}$

$I = \displaystyle\sf \dfrac{1}{2} \int \left( \dfrac{1}{ \sqrt{t }} + {t}^{ \frac{3}{2}} \right) .dt$

By using formula:-

→ $\displaystyle\sf \int {x}^{n} .dx = \dfrac{ {x}^{n + 1}}{n + 1}$

Therefore, $I = \sf \dfrac{1}{2} \left( \dfrac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } + \dfrac{{t}^{ \frac{5}{2}}}{ \frac{5}{2} } \right) + k$

$I = \sf \dfrac{1}{2} \left( 2 \times { {t}^{ \frac{1}{2} } } + {t}^{ \frac{5}{2}} \times { \frac{2}{5} } \right) + k$

$I = \sf {t}^{ \frac{1}{2} } + \frac{1}{5} {t}^{ \frac{5}{2} } + k$

Compare the equations:-

$\: \: \: \: { \huge{.}} \sf \: \: a = \dfrac{1}{2}$

$\: \: \: \: { \huge{.}} \sf \: \: b = \dfrac{5}{2}$

$\: \: \: \: { \huge{.}} \sf \: \: c = \dfrac{1}{5}$

=> $\sf a + b + c = \dfrac{1}{2} + \dfrac{5}{2} +\dfrac{1}{5}$

=> $\sf a + b + c =3 + \dfrac{1}{5}$

=> $\sf a + b + c = \dfrac{15 + 1}{5}$

=> $\sf a + b + c = \dfrac{16}{5}$