NIMCET 2023 — Mathematics PYQ

To solve this, we first simplify the integrand $\int x\sin x{\sec }^{3}xdx$.

Step 1: Simplify the integrand

We know that $\sec x=\frac{1}{\cos x}$, so $\sin x{\sec }^{3}x=\frac{\sin x}{{\cos }^{3}x}=\tan x{\sec }^{2}x$.

Therefore, the integral becomes:

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

Step 2: Apply Integration by Parts

We use the integration by parts formula: $\int udv=uv-\int vdu$.

Let $u=x⟹du=dx$

Let $dv=\tan x{\sec }^{2}xdx⟹v=\int \tan x{\sec }^{2}xdx=\frac{{\tan }^{2}x}{2}$

Applying the formula:

$$I=x\left(\frac{{\tan }^{2}x}{2}\right)-\int \frac{{\tan }^{2}x}{2}dx$$

$$I=\frac{x{\tan }^{2}x}{2}-\frac{1}{2}\int ({\sec }^{2}x-1)dx$$

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

$$I=\frac{1}{2}[x{\tan }^{2}x-\tan x+x]+c$$

Step 3: Rearrange to match the given form

Substitute ${\tan }^{2}x={\sec }^{2}x-1$:

$$I=\frac{1}{2}[x({\sec }^{2}x-1)-\tan x+x]+c$$

$$I=\frac{1}{2}[x{\sec }^{2}x-x-\tan x+x]+c$$

$$I=\frac{1}{2}[x{\sec }^{2}x-\tan x]+c$$

Now, we need to match this to the form $\frac{1}{2}[f(x){\sec }^{2}x+g(x)(\frac{\tan x}{x})]$.

We can write $-\tan x$ as $-x\cdot (\frac{\tan x}{x})$:

$$I=\frac{1}{2}[x{\sec }^{2}x+(-x)\left(\frac{\tan x}{x}\right)]+c$$

By comparing this with the required form:

• $f(x)=x$

• $g(x)=-x$

Step 4: Conclusion

Calculate $f(x)+g(x)$:

$$f(x)+g(x)=x+(-x)=0$$

Therefore, the correct option is C.