MHT CET · Maths · Indefinite Integration
The value of \(\int \frac{\sec x \cdot \tan x}{9-16 \tan ^2 x} \mathrm{~d} x\) is equal to
- A \(\frac{1}{24} \log \left(\frac{5+4 \sec x}{5-4 \sec x}\right)+\mathrm{c}\), (where c is a constant of integration)
- B \(\frac{1}{40} \log \left(\frac{5+4 \sec x}{5-4 \sec x}\right)+\mathrm{c}\), (where c is a constant of integration)
- C \(\frac{1}{24} \log \left(\frac{5-4 \sec x}{5+4 \sec x}\right)+\mathrm{c}\), (where c is a constant of integration)
- D \(\frac{1}{40} \log \left(\frac{5-4 \sec x}{5+4 \sec x}\right)+\mathrm{c}\), (where c is a constant of integration)
Answer & Solution
Correct Answer
(B) \(\frac{1}{40} \log \left(\frac{5+4 \sec x}{5-4 \sec x}\right)+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
Let I
\(\begin{aligned}
\mathrm{I} & =\int \frac{\sec x \tan x}{9-16 \tan ^2 x} \mathrm{~d} x \\
& =\int \frac{\sec x \tan x}{9-16\left(\sec ^2 x-1\right)} \mathrm{d} x \\
& =\int \frac{\sec x \tan x}{25-16 \sec ^2 x} \mathrm{~d} x
\end{aligned}\)
Put \(\sec x=\mathrm{t} \Rightarrow \sec x \tan x \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad \mathrm{I} & =\int \frac{\mathrm{dt}}{5^2-(4 \mathrm{t})^2} \\
& =\frac{1}{2(5)} \cdot \frac{1}{4} \log \left|\frac{5+4 \mathrm{t}}{5-4 t}\right| \\
& =\frac{1}{40} \log \left|\frac{5+4 \sec x}{5-4 \sec x}\right|+\mathrm{c}
\end{aligned}\)
\(\begin{aligned}
\mathrm{I} & =\int \frac{\sec x \tan x}{9-16 \tan ^2 x} \mathrm{~d} x \\
& =\int \frac{\sec x \tan x}{9-16\left(\sec ^2 x-1\right)} \mathrm{d} x \\
& =\int \frac{\sec x \tan x}{25-16 \sec ^2 x} \mathrm{~d} x
\end{aligned}\)
Put \(\sec x=\mathrm{t} \Rightarrow \sec x \tan x \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad \mathrm{I} & =\int \frac{\mathrm{dt}}{5^2-(4 \mathrm{t})^2} \\
& =\frac{1}{2(5)} \cdot \frac{1}{4} \log \left|\frac{5+4 \mathrm{t}}{5-4 t}\right| \\
& =\frac{1}{40} \log \left|\frac{5+4 \sec x}{5-4 \sec x}\right|+\mathrm{c}
\end{aligned}\)
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