Question

$\int \left({\sec }^2\right(2-3x)+\frac{1}{{\sin }^2(3-4x)}+\frac{\sin (2-3x)}{{\cos }^2(2-3x)}+\cos x{\text{cosec}}^{2}x+\text{cosec}(3x-2)\cot (3x-2))dx$

• A. $-\frac{1}{3}\tan (2-3x)+\frac{1}{4}[-\cot (3-4x\left)\right]-\frac{1}{3}\sec (2-3x)-\text{cosec}x-\frac{1}{3}\text{cosec}(3x-2).$
• B. $-\frac{1}{3}\tan (2-3x)-\frac{1}{4}[-\cot (3-4x\left)\right]-\frac{1}{3}\sec (2-3x)-\text{cosec}x-\frac{1}{3}\text{cosec}(3x-2).$
• C. $-\frac{1}{3}\tan (2-3x)-\frac{1}{4}[-\cot (3-4x\left)\right]-\frac{1}{3}\sec (2-3x)-\frac{1}{3}\text{cosec}(3x-2).$
• D. $-\frac{1}{3}\tan (2-3x)-\frac{1}{4}\left[\cot \right(3-4x\left)\right]-\frac{1}{3}\sec (2-3x)+\text{cosec}x-\frac{1}{3}\text{cosec}(3x-2).$

Answer 1

The correct option is B $-\frac{1}{3}\tan (2-3x)-\frac{1}{4}[-\cot (3-4x\left)\right]-\frac{1}{3}\sec (2-3x)-\text{cosec}x-\frac{1}{3}\text{cosec}(3x-2).$

$I=\int \left({\sec }^2\right(2-3x)+\frac{1}{{\sin }^2(3-4x)}+\frac{\sin (2-3x)}{{\cos }^2(2-3x)}+\cos x{\text{cosec}}^{2}x+\text{cosec}(3x-2\left)\cot \right(3x-2\left)\right)dx$

$\Rightarrow I=\int {\sec }^2(2-3x)dx+\int \frac{1}{{\sin }^2(3-4x)}dx+\int \frac{\sin (2-3x)}{{\cos }^2(2-3x)}dx$

$+\int \cos x{\text{cosec}}^{2}xdx+\int \text{cosec}(3x-2)\cot (3x-2)dx$

$\Rightarrow I=I_1+I_2+I_3+I_4+I_5$

Where

$I_1=\int {\sec }^2(2-3x)dx=-\frac{1}{3}\tan (2-3x)$

$I_2=\int \frac{1}{{\sin }^2(3-4x)}dx=\int {\csc }^2(3-4x)dx=-\frac{1}{4}(-\cot (3-4x\left)\right)$

$I_3=\int \frac{\sin (2-3x)}{{\cos }^2(2-3x)}dx=\int \tan (2-3x)\sec (2-3x)dx=-\frac{1}{3}\sec (2-3x)$

$I_4=\int \cos x{\text{cosec}}^{2}xdx=\int \cot x\csc xdx=-\csc x$

$I_5=\int \text{cosec}(3x-2)\cot (3x-2)dx=-\frac{1}{3}\text{cosec}(3x-2)$

Therefore

$I=-\frac{1}{3}\tan (2-3x)-\frac{1}{4}[-\cot (3-4x\left)\right]-\frac{1}{3}\sec (2-3x)$

$-\text{cosec}x-\frac{1}{3}\text{cosec}(3x-2)$

Hence, option 'B' is correct.