NIMCET 2007 — Mathematics PYQ

To solve an integral of the form $\int \frac{dx}{a\cos x+b\sin x}$, a standard technique is to multiply and divide the denominator by $\sqrt{a^2+b^2}$.

Here, the denominator is $1\cdot \cos x-1\cdot \sin x$. So, $a=1$ and $b=-1$.

$$\sqrt{a^2+b^2}=\sqrt{1^2+(-1{)}^2}=\sqrt{2}$$

Step 1: Transform the Denominator

Multiply and divide the expression inside the denominator by $\sqrt{2}$:

$$\cos x-\sin x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x\right)$$

We know from trigonometry values that $\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$ and $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$. Using the compound angle formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$:

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

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

Step 2: Substitute into the Integral

Now substitute this simplified form back into the main integral:

$$I=\int \frac{dx}{\sqrt{2}\sin \left(\frac{\pi }{4}-x\right)}$$

$$I=\frac{1}{\sqrt{2}}\int \csc \left(\frac{\pi }{4}-x\right)dx$$

Step 3: Apply the Cosecant Integration Formula

The standard formula for the integration of cosecant is $\int \csc \theta d\theta =\log \left|\tan \left(\frac{\theta }{2}\right)\right|$.

Remember to divide by the coefficient of $x$ (which is $-1$):

$$I=\frac{1}{\sqrt{2}}\cdot \frac{\log \left|\tan \left(\frac{\frac{\pi }{4}-x}{2}\right)\right|}{-1}+c$$

$$I=-\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{\pi }{8}-\frac{x}{2}\right)\right|+c$$

Step 4: Align with the Given Options

Since the options contain a positive $\frac{x}{2}$ term inside the tangent, we use the identity $\tan (-\theta )=-\tan \theta$ inside the modulus:

$$\left|\tan \left(\frac{\pi }{8}-\frac{x}{2}\right)\right|=\left|-\tan \left(\frac{x}{2}-\frac{\pi }{8}\right)\right|=\left|\tan \left(\frac{x}{2}-\frac{\pi }{8}\right)\right|$$

So the expression simplifies beautifully to:

$$I=-\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{x}{2}-\frac{\pi }{8}\right)\right|+c$$

To eliminate the negative sign outside the logarithm, we can use the property $-\log |A|=\log \left|\frac{1}{A}\right|=\log |\cot A|$.

$$I=\frac{1}{\sqrt{2}}\log \left|\cot \left(\frac{x}{2}-\frac{\pi }{8}\right)\right|+c$$

Alternatively, since $\cot (\theta )=\tan \left(\frac{\pi }{2}+\theta \right)$:

$$I=\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{\pi }{2}+\left(\frac{x}{2}-\frac{\pi }{8}\right)\right)\right|+c$$

$$I=\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{x}{2}+\frac{4\pi -\pi }{8}\right)\right|+c$$

$$I=\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{x}{2}+\frac{3\pi }{8}\right)\right|+c$$

Final Answer

The correct option is (d) $\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{x}{2}+\frac{3\pi }{8}\right)\right|+c$.