MHT CET · Maths · Definite Integration
\(\int \tan ^{-1}(\sec x+\tan x) d x=\)
- A \(\frac{\pi x}{4}+\frac{x^2}{4}+c\)
- B \(\sin x \cos x+c\)
- C \(\frac{\pi \mathrm{x}}{2}+\frac{\mathrm{x}^2}{2}+\mathrm{c}\)
- D \(\sin x+\cos x+c\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi x}{4}+\frac{x^2}{4}+c\)
Step-by-step Solution
Detailed explanation
\(\text {Let } I=\tan ^{-1}(\sec x+\tan x) d x \)
\( =\int \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x \)
\( =\int \tan ^{-1} x\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right]_{d x} \)
\( =\int \tan ^{-1}\left(\frac{\left.\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right) d x =\int \tan ^{-1}\left(\frac{1+\sin \frac{x}{2}}{1-\tan \frac{x}{2}}\right) d x \)
\( =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right] d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x \)
\( =\frac{\pi x}{4}+\frac{x^2}{4}+c\)
\( =\int \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x \)
\( =\int \tan ^{-1} x\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right]_{d x} \)
\( =\int \tan ^{-1}\left(\frac{\left.\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right) d x =\int \tan ^{-1}\left(\frac{1+\sin \frac{x}{2}}{1-\tan \frac{x}{2}}\right) d x \)
\( =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right] d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x \)
\( =\frac{\pi x}{4}+\frac{x^2}{4}+c\)
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