Q113 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics

$\begin{aligned} & \int \frac{d x}{(2 \sin x+\sec x)^4}=A(1+\tan x)^{-5} \\ & +B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k, \text { then } \\ & A+B+C=\end{aligned}$

  • A
    $\frac{-86}{105}$
  • B
    $\frac{-1}{105}$
  • C
    $\frac{-26}{105}$
  • Correct answer
    D
    $\frac{-16}{105}$

Solution

$\begin{aligned} & I=\int \frac{d x}{(2 \sin x+\sec x)^4}=\int \frac{\sec ^4 x}{\left(2 \tan x+\sec ^2 x\right)^4} d x \\ & =\int \frac{\left(1+\tan ^2 x\right)}{(1+\tan x)^8} \sec ^2 x d x\end{aligned}$ Let $\tan x=t \Rightarrow \sec ^2 x d x=d t$ So, $$ \begin{aligned} & I=\int \frac{1+t^2}{(1+t)^8} d t=\int \frac{(1+t)^2-2 t}{(1+t)^8} d t \\ &=\int \frac{d t}{(1+t)^6}-2 \int \frac{1+t-1}{(1+t)^8} d t \\ &=\int \frac{d t}{(1+t)^6}-2 \int \frac{d t}{(1+t)^7}+2 \int \frac{d t}{(1+t)^8} \\ &=-\frac{1}{5}(1+t)^{-5}+\frac{2}{6}(1+t)^{-6}-\frac{2}{7}(1+t)^{-7}+K \\ &=\frac{-1}{5}(1+\tan x)^{-5}+\frac{1}{3}(1+\tan x)^{-6} \\ &-\frac{2}{7}(1+\tan x)^{-7}+k \end{aligned} $$ So, $A=\frac{-1}{5}, B=\frac{1}{3}$ and $C=-\frac{2}{7}$ $$ \therefore A+B+C=-\frac{16}{105} \text {. } $$

Hence, the correct answer is option: 'D'.

Relevant Questions