NIMCET 2020 — Mathematics PYQ

Concept:

Derivatives of Trigonometric Functions:
- $\frac{d}{dx}\sin x=\cos x$
- $\frac{d}{dx}\cos x=-\sin x$
- $\frac{d}{dx}\tan x={\sec }^{2}x$
- $\frac{d}{dx}\cot x=-{\csc }^{2}x$
- $\frac{d}{dx}\sec x=\tan x\sec x$
- $\frac{d}{dx}\csc x=-\cot x\csc x$

Integration by Parts:

$\int f(x)g(x)dx=f(x)\int g(x)dx-\int [f^{\prime }(x)\int g(x)dx]dx.$

Calculation:
Integrating by parts by taking ${\csc }^{4}x$ as the first function and ${\sec }^{2}x$ as the second function:
$$\int {\sec }^{2}x{\csc }^{4}xdx$$
$$<br>={\csc }^{4}x\int {\sec }^{2}xdx-\int \left[\left(\frac{d}{dx}{\csc }^{4}x\right)\left(\int {\sec }^{2}xdx\right)\right]dx+C$$
$$={\csc }^{4}x\tan x-\int 4{\csc }^{3}x(-\cot x\csc x)\tan xdx+C$$
$$<br>={\csc }^{4}x\tan x+4\int {\csc }^{4}xdx+C$$
Integrating ${\csc }^{4}x$ by parts and taking ${\csc }^{2}x$ as the first and second functions:
$$\int ({\csc }^{2}x)({\csc }^{2}x)dx$$
$$<br>={\csc }^{2}x\int {\csc }^{2}xdx-\int \left[\left(\frac{d}{dx}{\csc }^{2}x\right)\left(\int {\csc }^{2}xdx\right)\right]dx+C$$
$$={\csc }^{2}x(-\cot x)-\int 2\csc x(-\cot x\csc x)(-\cot x)dx+C$$
$$<br>=-{\csc }^{2}x\cot x-2\int {\csc }^{2}x{\cot }^{2}xdx+C$$

Integrating csc${}^2$x cot${}^2$ x by parts and taking cot${}^2$ x as the first and csc${}^2$x as the second function:
$$\begin{array}{rl}& amp;\int {\csc }^2\mathbf{x}{\cot }^2\mathbf{x}\mathrm{d}\mathbf{x}\\ & amp;={\cot }^2\mathbf{x}\int {\csc }^2\mathbf{x}\mathrm{d}\mathbf{x}-\int \left[\left(\frac{\mathrm{d}}{\mathrm{d}\mathbf{x}}{\cot }^2\mathbf{x}\right)\left(\int {\csc }^2\mathbf{x}\mathrm{d}\mathbf{x}\right)\right]\mathrm{d}\mathbf{x}+\mathrm{C}\\ & amp;={\cot }^2\mathbf{x}(-\cot \mathbf{x})-\int 2\cot \mathbf{x}(-{\csc }^2\mathbf{x})(-\cot \mathbf{x})\mathrm{d}\mathbf{x}+\mathrm{C}\\ & amp;=-{\cot }^3\mathbf{x}-2\int {\csc }^2\mathbf{x}{\cot }^2\mathbf{x}\mathrm{d}\mathbf{x}+\mathrm{C}\\ & amp;\Rightarrow 3\int {\csc }^2\mathbf{x}{\cot }^2\mathbf{x}\mathrm{d}\mathbf{x}=-{\cot }^3\mathbf{x}+\mathrm{C}\\ & amp;\Rightarrow \int {\csc }^2\mathbf{x}{\cot }^2\mathbf{x}\mathrm{d}\mathbf{x}=-\frac{{\cot }^3\mathbf{x}}{3}+\mathrm{C}\\ & amp;\text{Finally,}\int {\sec }^2\mathbf{x}{\csc }^4\mathbf{x}\mathrm{d}\mathbf{x}\end{array}$$

$$={\csc }^{4}x\tan x+4\left[-{\csc }^{2}x\cot x-2\left(-\frac{{\cot }^{3}x}{3}\right)\right]+C$$
$$<br>={\csc }^{4}x\tan x-4{\csc }^{2}x\cot x+\frac{8}{3}{\cot }^{3}x+C$$
$$=(1+{\cot }^{2}x{)}^2\tan x-4(1+{\cot }^{2}x)\cot x+\frac{8}{3}{\cot }^{3}x+C$$
$$<br>=(1+2{\cot }^{2}x+{\cot }^{4}x)\tan x-4\cot x-4{\cot }^{3}x+\frac{8}{3}{\cot }^{3}x+C$$
$$=\tan x+2\cot x+{\cot }^{3}x-4\cot x-\frac{4}{3}{\cot }^{3}x+C$$
$$<br>=-\frac{1}{3}{\cot }^{3}x+\tan x-2\cot x+C$$
$$\therefore \text{Comparing }\left(-\frac{1}{3}{\cot }^{3}x+\tan x-2\cot x+C\right)\text{ with</span><span style="font-family: arial, helvetica, sans-serif;"> }</span><spanstyle="font-family:arial,helvetica,sans-serif;">⟨/span><spanstyle="font-family:arial,helvetica,sans-serif;">(-\frac{1}{3}{\cot }^{3}x+k\tan x-2\cot x+C)\text{ we can say that }k=1.$$