\int dx \over \cos x + \sqrt 3 \sin x equals

\,1 \over 2\,\log \,\tan \,( x \over 2 + \pi \over 12 ) + C

\log \,\tan \,( x \over 2 + \pi \over 12 ) + C

\log \,\tan \,( x \over 2 - \pi \over 12 ) + C

\,1 \over 2\,\log \,\tan \,( x \over 2 - \pi \over 12 ) + C

Indefinite Integrals PYQ — JEE Main 2007 · Shift 0 · Q50 with Solution

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Correct answer: C

Simplify the denominator

We use the identity $a\cos x+b\sin x=R\cos(x-\phi),$ where $R=\sqrt{a^2+b^2}.$

Here, $\cos x+\sqrt{3}\sin x.$ So, $R=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2.$

Let $\cos x+\sqrt{3}\sin x=2\cos(x-\phi).$ Then $2\cos(x-\phi)=2(\cos x\cos\phi+\sin x\sin\phi).$ Comparing coefficients, $2\cos\phi=1 \Rightarrow \cos\phi=\frac12,$ $2\sin\phi=\sqrt3 \Rightarrow \sin\phi=\frac{\sqrt3}{2}.$ Hence, $\phi=\frac{\pi}{3}.$ Therefore, $\cos x+\sqrt3\sin x=2\cos\left(x-\frac\pi3\right).$

So the integral becomes

=\frac12\int \sec\left(x-\frac\pi3\right)dx.$$ 2. **Integrate using the standard formula** We know $$\int \sec u\,du=\log\left|\sec u+\tan u\right|+C.$$ Thus, $$I=\frac12\log\left|\sec\left(x-\frac\pi3\right)+\tan\left(x-\frac\pi3\right)\right|+C.$$ Now use the identity $$\sec \theta+\tan \theta=\tan\left(\frac\pi4+\frac\theta2\right).$$ So, $$I=\frac12\log\tan\left(\frac\pi4+\frac12\left(x-\frac\pi3\right)\right)+C.$$ Simplifying, $$\frac\pi4+\frac{x}{2}-\frac\pi6 =\frac{x}{2}+\frac\pi{12}.$$ Hence, $$I=\frac12\log\tan\left(\frac{x}{2}+\frac\pi{12}\right)+C.$$ 3. **Match with the options** This is exactly **Option C**: $$\boxed{\frac12\log\tan\left(\frac{x}{2}+\frac\pi{12}\right)+C}.$$

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View written solutionFree

Correct answer: C

Simplify the denominator

We use the identity $a\cos x+b\sin x=R\cos(x-\phi),$ where $R=\sqrt{a^2+b^2}.$

Here, $\cos x+\sqrt{3}\sin x.$ So, $R=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2.$

So the integral becomes

=\frac12\int \sec\left(x-\frac\pi3\right)dx.$$ 2. **Integrate using the standard formula** We know $$\int \sec u\,du=\log\left|\sec u+\tan u\right|+C.$$ Thus, $$I=\frac12\log\left|\sec\left(x-\frac\pi3\right)+\tan\left(x-\frac\pi3\right)\right|+C.$$ Now use the identity $$\sec \theta+\tan \theta=\tan\left(\frac\pi4+\frac\theta2\right).$$ So, $$I=\frac12\log\tan\left(\frac\pi4+\frac12\left(x-\frac\pi3\right)\right)+C.$$ Simplifying, $$\frac\pi4+\frac{x}{2}-\frac\pi6 =\frac{x}{2}+\frac\pi{12}.$$ Hence, $$I=\frac12\log\tan\left(\frac{x}{2}+\frac\pi{12}\right)+C.$$ 3. **Match with the options** This is exactly **Option C**: $$\boxed{\frac12\log\tan\left(\frac{x}{2}+\frac\pi{12}\right)+C}.$$

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