NIMCET 2020 — Mathematics PYQ

Step 1: Simplify the given integral.

Let the given integral be:

$$I=\int {\sec }^{2}x{\csc }^{4}xdx$$

We know the trigonometric identity ${\sin }^{2}x+{\cos }^{2}x=1$. Let's substitute $1$ in the numerator as $({\sin }^{2}x+{\cos }^{2}x{)}^2=1$:

$$I=\int \frac{1}{{\cos }^{2}x\cdot {\sin }^{4}x}dx$$

Substitute $1={\sin }^{2}x+{\cos }^{2}x$:

$$I=\int \frac{({\sin }^{2}x+{\cos }^{2}x)}{{\cos }^{2}x\cdot {\sin }^{4}x}dx$$

$$I=\int \left(\frac{{\sin }^{2}x}{{\cos }^{2}x{\sin }^{4}x}+\frac{{\cos }^{2}x}{{\cos }^{2}x{\sin }^{4}x}\right)dx$$

$$I=\int \left(\frac{1}{{\cos }^{2}x{\sin }^{2}x}+\frac{1}{{\sin }^{4}x}\right)dx$$

Step 2: Express everything in terms of $\tan x$ or $\cot x$.

Divide the numerator and denominator of the first term by ${\cos }^{4}x$, or simply substitute $1={\sin }^{2}x+{\cos }^{2}x$ again in the first term:

$$I=\int \left(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x{\sin }^{2}x}+{\csc }^{4}x\right)dx$$

$$I=\int \left(\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}+{\csc }^{4}x\right)dx$$

$$I=\int \left({\sec }^{2}x+{\csc }^{2}x+{\csc }^{4}x\right)dx$$

Using the identity ${\csc }^{2}x=1+{\cot }^{2}x$, rewrite ${\csc }^{4}x$:

$${\csc }^{4}x={\csc }^{2}x\cdot {\csc }^{2}x=(1+{\cot }^{2}x){\csc }^{2}x$$

Now, substitute this back into the integral:

$$I=\int \left({\sec }^{2}x+{\csc }^{2}x+(1+{\cot }^{2}x){\csc }^{2}x\right)dx$$

$$I=\int \left({\sec }^{2}x+2{\csc }^{2}x+{\cot }^{2}x{\csc }^{2}x\right)dx$$

Step 3: Integrate each term.

We know the standard integration formulas:

• $\int {\sec }^{2}xdx=\tan x$

• $\int {\csc }^{2}xdx=-\cot x$

For the third term $\int {\cot }^{2}x{\csc }^{2}xdx$, let $u=\cot x$, then $du=-{\csc }^{2}xdx$:

$$\int {\cot }^{2}x{\csc }^{2}xdx=\int u^2(-du)=-\frac{u^3}{3}=-\frac{1}{3}{\cot }^{3}x$$

Combining all the integrals together:

$$I=\tan x+2(-\cot x)-\frac{1}{3}{\cot }^{3}x+C$$

$$I=-\frac{1}{3}{\cot }^{3}x+1\tan x-2\cot x+C$$

Step 4: Compare with the given equation.

The question gives the result as:

$$I=-\frac{1}{3}{\cot }^{3}x+k\tan x-2\cot x+C$$

By comparing the coefficients of $\tan x$, we get:

$$k=1$$

Correct Option: A (1)