6. ⇒ (MHT CET 2023 9th May Evening Shift )

$\int (\sqrt{\tan x} + \sqrt{\cot x}) d x =$

A. $\sqrt{2} \sin^{- 1} (\sin x - \cos x) + c$ , where $c$ is a constant of integration.

B. $\frac{1}{\sqrt{2}} \sin^{- 1} (\sin x - \cos x) + c$ , where $c$ is a constant of integration.

C. $\sin^{- 1} (\sin x - \cos x) + c$ , where c is a constant of integration.

D. $2 \sin^{- 1} (\sin x - \cos x) + c$ , where $c$ is a constant of integration.

Correct Option is (A)

$\begin{aligned} Let I = \int (\sqrt{\tan x} + \sqrt{\cot x}) d x \\ = \int (\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}) d x \\ I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x \\ Let \sin x - \cos x = t \\ ∴ (\sin x + \cos x) d x = dt \\ Consider, \sin x - \cos x = t \\ Squaring on both sides, we get \\ 1 - 2 \sin t \cdot \cos t = t^{2} \\ 1 - t^{2} = 2 \sin t \cdot \cos t \\ ∴ \sin t \cdot \cos t = \frac{1 - t^{2}}{2} \\ ∴ I = \int \frac{dt}{\frac{\sqrt{1 - t^{2}}}{\sqrt{2}}} \\ ∴ I = \sqrt{2} \int \frac{1}{\sqrt{1 - t^{2}}} dt \\ ∴ I = \sqrt{2} \sin^{- 1} (t) + c \\ ∴ I = \sqrt{2} \sin^{- 1} (\sin x - \cos x) + c \end{aligned}$

7. ⇒ (MHT CET 2023 9th May Morning Shift )

If $I = \int \frac{\sin x + \sin^{3} x}{\cos 2 x} d x = P \cos x + Q \log | \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} |$ (where $c$ is a constant of integration), then values of $P$ and $Q$ are respectively

A. $\frac{1}{2}, \frac{3}{4 \sqrt{2}}$

B. $\frac{1}{2}, \frac{- 3}{4 \sqrt{2}}$

C. $\frac{1}{2}, \frac{3}{2 \sqrt{2}}$

D. $\frac{1}{2}, \frac{- 3}{2 \sqrt{2}}$

Correct Option is (B)

$\begin{aligned} I & = \int \frac{\sin x + \sin^{3} x}{\cos 2 x} d x \\ = \int \frac{\sin x (1 + \sin^{2} x)}{2 \cos^{2} x - 1} d x \\ = \int \frac{\sin x (2 - \cos^{2} x)}{2 \cos^{2} x - 1} d x \end{aligned}$

$\begin{aligned} Put \cos x = t \\ - \sin x d x = dt \\ I = \int \frac{t^{2} - 2}{2 t^{2} - 1} d t \\ = \frac{1}{2} \int \frac{2 t^{2} - 4}{2 t^{2} - 1} d t \\ = \frac{1}{2} [\int \frac{2 t^{2} - 1}{2 t^{2} - 1} d t - \int \frac{3}{2 t^{2} - 1} d t] \\ I = \frac{1}{2} t - \frac{1}{2} \times \frac{3}{2 \sqrt{2}} \log | \frac{\sqrt{2} t - 1}{\sqrt{2} t + 1} | + c \\ = \frac{1}{2} \cos x - \frac{3}{4 \sqrt{2}} \log | \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} | + c \\ ∴ P = \frac{1}{2} and Q = \frac{- 3}{4 \sqrt{2}} \end{aligned}$

8. ⇒ (MHT CET 2021 20th September Evening Shift )

$\int \frac{2 x^{2} - 1}{x^{4} - x^{2} - 20} d x =$

A. $\frac{1}{\sqrt{5}} \log | \frac{x + \sqrt{5}}{x - \sqrt{5}} | + \tan^{- 1} (\frac{x}{2}) + c$

B. $\frac{1}{2 \sqrt{5}} \log | \frac{x + \sqrt{5}}{x - \sqrt{5}} | + \tan^{- 1} (\frac{x}{2}) + c$

C. $\frac{1}{2 \sqrt{5}} \log | \frac{x - \sqrt{5}}{x + \sqrt{5}} | + \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + c$

D. $\frac{1}{2} \log | \frac{x - \sqrt{5}}{x + \sqrt{5}} | + \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + c$

Correct Option is (C)

Let $I = \int \frac{2 x^{2} - 1}{x^{4} - x^{2} - 20} d x$

If $x^{2} = t$ , then $\frac{2 x^{2} - 1}{x^{4} - x^{2} - 20} = \frac{2 t - 1}{t^{2} - t - 20}$

Let $\frac{2 t - 1}{(t - 5) (t + 4)} = \frac{A}{(t - 5)} + \frac{B}{(t + 4)}$

$\begin{aligned} ∴ 2 t - 1 = (t + 4) A + (t - 5) B \\ ∴ 2 = A + B and - 1 = 4 A - 5 B \end{aligned}$

Solving, we get $B = 1, A = 1$

$∴ I = \int [\frac{1}{x^{2} - 5} + \frac{1}{x^{2} + 4}] d x = \frac{1}{2 \sqrt{5}} \log | \frac{x - \sqrt{5}}{x + \sqrt{5}} | + \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + c$

⇒Indefinite Integration