Definite Integral

Q: If value of $\displaystyle \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0) $ the values of A and B are

(a) $\displaystyle A =\frac{\pi}{2} , B = 0 $

(b) $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha} $

(c) $\displaystyle A =\frac{\pi}{6} , B = \frac{\pi}{ sin\alpha} $

(d) $\displaystyle A =\pi , B = \frac{\pi}{ sin\alpha} $

Ans: (a) , (b)

Sol: $\displaystyle I = \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} $

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{cos^2 \frac{x}{2} + sin^2 \frac{x}{2}-cos\alpha (cos^2 \frac{x}{2} – sin^2 \frac{x}{2})} $

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{(1-cos\alpha) cos^2 \frac{x}{2} + (1 + cos\alpha) sin^2 \frac{x}{2}} $

$\displaystyle I = \int_{0}^{\alpha} \frac{sec^2 \frac{x}{2} dx}{2 sin^2 \frac{\alpha}{2} + 2 cos^2 \frac{\alpha}{2} tan^2 \frac{x}{2}} $

$\displaystyle I = \frac{1}{2} \int_{0}^{\alpha} \frac{sec^2 \frac{\alpha}{2} sec^2 \frac{x}{2} dx}{tan^2 \frac{\alpha}{2} + tan^2 \frac{x}{2}} $

Let $tan\frac{x}{2} = t $

$\displaystyle I = \int_{0}^{tan(\alpha/2)} \frac{sec^2 \frac{\alpha}{2} dt }{t^2 + tan^2 \frac{\alpha}{2}} $

$\displaystyle I = sec^2 \frac{\alpha}{2} cot\frac{\alpha}{2} [tan^{-1}\frac{t}{tan(\alpha/2)}]_{0}^{tan(\alpha/2)} $

$\displaystyle = \frac{2}{sin\alpha}.\frac{\pi}{4}= \frac{\pi}{2 sin\alpha}$

$\displaystyle A =\frac{\pi}{2} , B = 0 $

Also , $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha} $

Hence (a), (b) are correct answers.

Q: The value of $\displaystyle \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $ is

(A) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a > 0 , b > 0$

(B) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a < 0 , b < 0$

(C) $\displaystyle \frac{\pi}{4} ; a = 1 , b = 1 $

(D) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{a}{b}) + \frac{1}{ab}$

Ans: (A) , (B) , (C)

Sol: $\displaystyle I = \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $

Dividing N^r and D^r by cos²x ,

$\displaystyle I = \int_{0}^{\pi/4} \frac{sec^2 x dx}{a^2 + b^2 tan^2 x} $

Put tanx = t ; sec²x dx = dt

$\displaystyle I = \int_{0}^{1} \frac{dt}{a^2 + b^2 t^2} $

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + \frac{a^2}{b^2}} $

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + (\frac{a}{b})^2} $

$\displaystyle = \frac{b}{a b^2} [tan^{-1}\frac{b t}{a}]_{0}^{1} $

$\displaystyle = \frac{1}{a b} [tan^{-1}\frac{b}{a}] $ ; (a ≠0 , b≠ 0)

Also for a = 1, b = 1

$\displaystyle I = \int_{0}^{\pi/4}dx = \frac{\pi}{4} $

Hence (A), (B), (C) are correct answers.

Q: If $\displaystyle f(x) = \int_{x}^{x^2} \frac{dt}{(logt)^2} , x \ne 0$ then f(x) is

(A) monotonically increasing in (2 , ∞)

(B) monotonically decreasing in (1 , 2)

(C) monotonically decreasing in (2 , ∞)

(D) monotonically decreasing in (0 , 1)

Ans: (A) , (B)

Sol: $\displaystyle f'(x) = \frac{2x}{(logx^2)^2} – \frac{1}{(logx)^2}$

$\displaystyle f'(x) = \frac{x-2}{2(logx)^2} $

f'(x) > 0 for x > 2

and f'(x) < 0 for x < 2

Hence (A), (B) are correct answers.

Q: $\displaystyle \int_{0}^{\pi} x f(sin x) dx $ is equal to

(A) $ \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx $

(B) $ \pi \int_{0}^{\pi/2} f(sin x) dx $

(C) $ 2 \pi \int_{0}^{\pi/2} f(sin x) dx $

(D) none of these

Ans: (A) , (B)

Sol: Let $\displaystyle I = \int_{0}^{\pi} x f(sin x) dx $

$\displaystyle = \int_{0}^{\pi} (\pi – x ) f sin (\pi-x) dx $

$\displaystyle I = \pi \int_{0}^{\pi} f(sin x) dx – I $

$\displaystyle 2 I = \pi \int_{0}^{\pi} f(sin x) dx $

$\displaystyle I = \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx $

$\displaystyle I = 2 \times \frac{\pi}{2} \int_{0}^{\pi/2} f(sin x) dx $

$\displaystyle I = \pi \int_{0}^{\pi/2} f(sin x) dx $

Hence (A) and (B) are the correct answers.

Q: If $\displaystyle I = \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 + sin^3 x}} $ , then

(A) 0 < I < 1

(B) $ I > \frac{\pi}{2\sqrt{2}}$

(C) I < √2 π

(D) I > 2 π

Ans: (B) , (C)

Sol: Since x ∈ [0 , π/2 ]

⇒ 1 ≤ 1 + sin³ x  ≤  2

$\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{1 + sin^3 x}} \le 1 $

$ \int_{0}^{\pi/2} \frac{1}{\sqrt{2}} dx \le \int_{0}^{\pi/2} \frac{1}{\sqrt{1 + sin^3 x}} dx \le \int_{0}^{\pi/2} 1 dx $

$\frac{\pi}{2 \sqrt{2}} \le I \le \frac{\pi}{2}$

Hence (B) and (C) are the correct answers.