Question

If $\displaystyle \sf \int \dfrac{dx}{(2sin \ x + sec \ x)^4} = A(1 + tan \ x)^{-5} + B(1 + tan \ x)^{-6} + C(1 + tan \ x)^{-7} + k$, then A + B + C is :
a) -86/105
b) -1/105
c) -26/105
d) -16/105​

Answer 1

Let us evaluate,

$\displaystyle\sf{\longrightarrow I=\int\dfrac{1}{(2\sin x+\sec x)^4}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{1}{\left(2\sin x+\dfrac{1}{\cos x}\right)^4}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{\cos^4x}{\left(1+\sin(2x)\right)^4}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{\cos^4x}{\left(\left(\sin x+\cos x\right)^2\right)^4}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{\cos^4x}{\left(\sin x+\cos x\right)^8}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{\dfrac{\cos^4x}{\cos^8x}}{\left(\dfrac{\sin x+\cos x}{\cos x}\right)^8}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{\sec^4x}{\left(1+\tan x\right)^8}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{\sec^2x\cdot\sec^2x}{\left(1+\tan x\right)^8}\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{1+\tan^2x}{\left(1+\tan x\right)^8}\cdot\sec^2x\ dx}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{(1+\tan x)^2-2\tan x}{\left(1+\tan x\right)^8}\cdot\sec^2x\ dx}$

Substitute $\sf{u=1+\tan x.}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{u^2-2(u-1)}{u^8}\ du}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{u^2-2u+2}{u^8}\ du}$

$\displaystyle\sf{\longrightarrow I=\int\dfrac{1}{u^6}\ du-2\int\dfrac{1}{u^7}\ du+2\int\dfrac{1}{u^8}\ du}$

$\displaystyle\sf{\longrightarrow I=\dfrac{u^{-5}}{-5}-2\cdot\dfrac{u^{-6}}{-6}+2\cdot\dfrac{u^{-7}}{-7}+k}$

Undoing the substitution,

$\displaystyle\sf{\longrightarrow I=-\dfrac{1}{5}(1+\tan x)^{-5}+\dfrac{1}{3}(1+\tan x)^{-6}-\dfrac{2}{7}(1+\tan x)^{-7}+k\quad\quad\dots(1)}$

But in the question it's given,

$\displaystyle\sf{\longrightarrow I=A(1+\tan x)^{-5}+B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k\quad\quad\dots(2)}$

Comparing (1) and (2) we get,

• $\sf{A=-\dfrac{1}{5}}$

• $\sf{B=\dfrac{1}{3}}$

• $\sf{C=-\dfrac{2}{7}}$

Therefore,

$\sf{\longrightarrow A+B+C=-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{2}{7}}$

$\sf{\longrightarrow\underline{\underline{A+B+C=-\dfrac{16}{105}}}}$

Hence (D) is the answer.

Answer 2

GIVEN :–

$\\ \implies \displaystyle \sf \int \dfrac{dx}{(2sin \ x + sec \ x)^4} = A(1 + tan \ x)^{-5} + B(1 + tan \ x)^{-6} + C(1 + tan \ x)^{-7} + k$

TO FIND :–

• A + B + C = ?

SOLUTION :–

• Let –

$\\ \displaystyle \implies \bf I = \int \dfrac{dx}{(2sin \ x + sec \ x)^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{dx}{ \bigg(2sin \ x + \dfrac{1}{ \cos x} \bigg)^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{(2sin \ x \cos x + 1)^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{(2sin \ x \cos x + { \sin}^{2}x + { \cos}^{2} x )^4}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{({ \sin}x + { \cos}x )^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \cos}^{4}x. dx}{ \cos^{8} (x) \bigg( \dfrac{{ \sin}(x)}{ \cos(x) } + 1 \bigg )^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{dx}{ \cos^{4} (x) ( \tan x+ 1)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{\sec^{4} (x).dx}{ ( \tan x+ 1)^8}$

• We should write this as –

$\\ \displaystyle \implies \bf I = \int \dfrac{ { \sec}^{2}(x). \sec^{2} (x).dx}{ ( \tan x+ 1)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + { \tan}^{2}(x) \} \sec^{2} (x).dx}{ ( \tan x+ 1)^8}$

• Put 1 + tan(x) = t –

• So that –

$\\ \displaystyle \implies \bf { \sec}^{2}(x).dx = dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + {t}^{2}\}. dt}{ ( 1+ t)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ \{ 1 + {(t - 1)}^{2}\}. dt}{ (t)^8}$

$\\ \displaystyle \implies \bf I = \int \dfrac{ 1 + {(t - 1)}^{2}}{ (t)^8} .dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{ 1 + {t}^{2} + 1 - 2t }{ (t)^8} .dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{ {t}^{2} - 2t + 2 }{ (t)^8} .dt$

$\\ \displaystyle \implies \bf I = \int \dfrac{1}{ {t}^{6} }dt - 2 \int \dfrac{1}{ {t}^{7} } dt + 2 \int \dfrac{1}{ {t}^{8} } dt$

$\\ \displaystyle \implies \bf I = - \dfrac{1}{5 {t}^{5} } + (2) \dfrac{1}{6{t}^{6} } - (2 ) \dfrac{1}{7 {t}^{7} } + k$

• Now replace 't' –

$\\ \displaystyle \implies \bf I = - \dfrac{ {(1 + \tan x) }^{ - 5} }{5} + \dfrac{{(1 + \tan x) }^{ -6}}{3} - \dfrac{2{(1 + \tan x) }^{ - 7}}{7} + k$

• Now compare –

$\\ \implies \bf A = - \dfrac{ 1}{5}$

$\\ \implies \bf B = \dfrac{ 1}{3}$

$\\ \implies \bf C = - \dfrac{2}{7}$

• So that –

$\\ \implies \bf A+ B+ C =- \dfrac{ 1}{5} + \dfrac{ 1}{3} - \dfrac{2}{7}$

$\\ \implies \bf A+ B+ C =- \dfrac{16}{105}$

• Hence Option (D) is correct.