If ∫ (a x-2 x)/(b x+5 x) d x=(7)/(41) x+(22)/(7) |b x+5 x|+C,(a>0, b>0) , then ∫ (d x)/(b+a x)=

Correct answer: B. 2 7 ^ -1 ( (x)/(2) 7 )+C We have, ∫ (a x-2 x)/(b x+5 x) d x=(7)/(41) x+(22)/(41) aligned & |(b x+5 x)|+C & Now, let a x-2 x=λ(b x+5 x) & +μ(b x-5 x) aligned a=5 λ+μ b and -2=b λ-5 μ But it is given, λ=(7)/(41) and μ=(22)/(41)a=3, b=4∫ (d x)/(b+a x)=∫ (d x)/(4+3 x)=∫ ((1+ ^2 (x)/(2)) d x)/(4(1+ ^2 (x)/(2))+3(1- ^2 (x)/(2))) aligned &=∫ ( ^2 x / 2)/(7+ ^2 x / 2) d x & (x)/(2)=t ^2 (x)/(2) d x=2 d t & (d x)/(b+a x)=2 ∫ (d t)/(7+t^2) &= 2 7 ^ -1 t 7 +C &= 2 7 ^ -1 ( x / 2 7 )+C aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main · Mathematics · Indefinite Integration

If ∫ (a x-2 x)/(b x+5 x) d x=(7)/(41) x+(22)/(7) |b x+5 x|+C,(a>0,...

\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{7} \log|b \sin x+5 \cos x|+C,(a>0, b>0)

, then

\int \frac{d x}{b+a \cos x}=

Options

A $\frac{2}{3} \log \left(\frac{3 \tan \frac{x}{2}+4-\sqrt{3}}{3 \tan \frac{x}{2}+4+\sqrt{3}}\right)+C$
B $\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{7}}\right)+C$
C $\frac{2}{\sqrt{7}} \log \left(\frac{\sqrt{7}-\tan \frac{x}{2}}{\sqrt{7}+\tan \frac{x}{2}}\right)+C$
D $2 \sinh ^{-1}\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+C$

Correct answer

B. $\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{7}}\right)+C$

Step-by-step solution

We have,

\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41}

\begin{aligned} & \log |(b \sin x+5 \cos x)|+C \\ & \text { Now, let } a \cos x-2 \sin x=\lambda(b \sin x+5 \cos x) \\ & \qquad+\mu(b \cos x-5 \sin x)\end{aligned}

\therefore a=5 \lambda+\mu b

and

-2=b \lambda-5 \mu

But it is given,

\lambda=\frac{7}{41}

and

\mu=\frac{22}{41}a=3, b=4\int \frac{d x}{b+a \cos x}=\int \frac{d x}{4+3 \cos x}=\int \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{4\left(1+\tan ^2 \frac{x}{2}\right)+3\left(1-\tan ^2 \frac{x}{2}\right)}

\begin{aligned} &=\int \frac{\sec ^2 x / 2}{7+\tan ^2 x / 2} d x \\ & \tan \frac{x}{2}=t \Rightarrow \sec ^2 \frac{x}{2} d x=2 d t \\ & \frac{d x}{b+a \cos x}=2 \int \frac{d t}{7+t^2} \\ &=\frac{2}{\sqrt{7}} \tan ^{-1} \frac{t}{\sqrt{7}}+C \\ &=\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\tan x / 2}{\sqrt{7}}\right)+C\end{aligned}

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