\int dx \over \cos x - \sin x is equal to

1 \over \sqrt 2 \log | \tan ( x \over 2 + 3\pi \over 8 ) | + C

1 \over \sqrt 2 \log | \cot ( x \over 2 ) | + C

1 \over \sqrt 2 \log | \tan ( x \over 2 - 3\pi \over 8 ) | + C

\,1 \over \sqrt 2 \log | \tan ( x \over 2 - \pi \over 8 ) | + C

Indefinite Integrals PYQ — JEE Main 2004 · Shift 0 · Q83 with Solution

View written solutionFree

Correct answer: A

Rewrite the denominator

We use the identity

\cos x - \sin x = \sqrt{2}\cos\left(x+\frac{\pi}{4}\right).

So,

\int \frac{dx}{\cos x-\sin x} = \frac{1}{\sqrt{2}}\int \sec\left(x+\frac{\pi}{4}\right)\,dx.

Use the standard integral

Recall:

\int \sec u\,du = \ln|\sec u+\tan u|+C.

Let

u = x+\frac{\pi}{4}, \quad d\nu = dx.

Then

\int \frac{dx}{\cos x-\sin x} = \frac{1}{\sqrt{2}}\ln\left|\sec\nu+\tan\nu\right|+C.

Substituting back,

= \frac{1}{\sqrt{2}}\ln\left|\sec\left(x+\frac{\pi}{4}\right)+\tan\left(x+\frac{\pi}{4}\right)\right|+C.

Convert to half-angle form

Use the identity

\sec \theta + \tan \theta = \tan\left(\frac{\theta}{2}+\frac{\pi}{4}\right).

Taking

\theta = x+\frac{\pi}{4},

we get

\sec\left(x+\frac{\pi}{4}\right)+\tan\left(x+\frac{\pi}{4}\right) = \tan\left(\frac{x+\pi/4}{2}+\frac{\pi}{4}\right).

Now simplify the angle:

\frac{x+\pi/4}{2}+\frac{\pi}{4} = \frac{x}{2}+\frac{\pi}{8}+\frac{\pi}{4} = \frac{x}{2}+\frac{3\pi}{8}.

Hence,

\int \frac{dx}{\cos x-\sin x} = \frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}+\frac{3\pi}{8}\right)\right|+C.

Match with options

This matches Option A.

Final Answer:

\boxed{\frac{1}{\sqrt 2 }\log \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + C}

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View written solutionFree

Correct answer: A

Rewrite the denominator

We use the identity

\cos x - \sin x = \sqrt{2}\cos\left(x+\frac{\pi}{4}\right).

So,

\int \frac{dx}{\cos x-\sin x} = \frac{1}{\sqrt{2}}\int \sec\left(x+\frac{\pi}{4}\right)\,dx.

Use the standard integral

Recall:

\int \sec u\,du = \ln|\sec u+\tan u|+C.

Let

u = x+\frac{\pi}{4}, \quad d\nu = dx.

Then

\int \frac{dx}{\cos x-\sin x} = \frac{1}{\sqrt{2}}\ln\left|\sec\nu+\tan\nu\right|+C.

Substituting back,

= \frac{1}{\sqrt{2}}\ln\left|\sec\left(x+\frac{\pi}{4}\right)+\tan\left(x+\frac{\pi}{4}\right)\right|+C.

Convert to half-angle form

Use the identity

\sec \theta + \tan \theta = \tan\left(\frac{\theta}{2}+\frac{\pi}{4}\right).

Taking

\theta = x+\frac{\pi}{4},

we get

\sec\left(x+\frac{\pi}{4}\right)+\tan\left(x+\frac{\pi}{4}\right) = \tan\left(\frac{x+\pi/4}{2}+\frac{\pi}{4}\right).

Now simplify the angle:

\frac{x+\pi/4}{2}+\frac{\pi}{4} = \frac{x}{2}+\frac{\pi}{8}+\frac{\pi}{4} = \frac{x}{2}+\frac{3\pi}{8}.

Hence,

\int \frac{dx}{\cos x-\sin x} = \frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}+\frac{3\pi}{8}\right)\right|+C.

Match with options

This matches Option A.

Final Answer:

\boxed{\frac{1}{\sqrt 2 }\log \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + C}

Indefinite Integrals question

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