Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type to help strengthen their preparations for exams.
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type
Question 1.
Evaluate \(\int_0^{\pi / 2} \frac{d x}{4+5 \cos x}\). [(TS) May ’18; (AP) Mar. ’16, ’15]
Solution:
Put tan \(\frac{x}{2}\) = t then \(\sec ^2 \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}\)
Question 2.
Evaluate \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\). [(AP) Mar. ’17; (TS) May ’15]
Solution:
Put sin x – cos x = t
⇒ (sin x – cos x)2 = t2
(cos x + sin x ) dx = dt
sin2x + cos2x – 2 sin x cos x = t2
1 – sin 2x = t2
sin 2x = 1 – t2
Lower limit: x = 0 ⇒ t = -1
Upper limit: x = \(\frac{\pi}{4}\) ⇒ t = 0
Question 3.
Evaluate \(\int_0^1 \frac{\log (1+x)}{1+x^2} d x\). [(TS) Mar. ’20; (TS) May ’19, ’17; ’12]
Solution:
Put x = tan θ
then dx = sec2θ dθ
Lower limit: x = 0 ⇒ θ = 0
Upper limit: x = 1 ⇒ θ = \(\frac{\pi}{4}\)
Question 4.
Show that \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x=\frac{\pi}{2 \sqrt{2}} \log (\sqrt{2}+1)\). [(AP) Mar. ’20, ’18; ’17 (TS)]
Solution:
Question 5.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^2 x}{\cos x+\sin x} d x\). [(TS) May ’15]
Solution:
Question 6.
Evaluate \(\int_0^\pi \frac{x}{1+\sin x} d x\)
Solution:
Question 7.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\sin x} d x\). [Mar. ’16 (TS); May; Mar. ’15 (AP); March ’13]
Solution:
Question 8.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\). [(AP) May ’18, ’16, ’14]
Solution:
Question 9.
Evaluate \(\frac{x \sin ^3 x}{1+\cos ^2 x} d x\). [(TS) May ’18; Mar. ’15]
Solution:
Question 10.
Evaluate \(\int_3^7 \sqrt{\frac{7-x}{x-3}} d x\)
Solution:
Put x = 3 cos2θ + 7 sin2θ then
dx = (3 . 2 cos θ (-sin θ) + 7 . 2 sin θ cos θ) dθ
= (-3 . 2 sin θ . cos θ + 7 . 2 sin θ cos θ) dθ
= 4 . 2 sin θ cos θ dθ
= 4 sin 2θ dθ
7 – x = 7 – 3 cos2θ – 7 sin2θ
= 7(1- sin2θ) – 3 cos2θ
= 7 cos2θ – 3 cos2θ
= 4 cos2θ
x – 3 = 3 cos2θ + 7 sin2θ – 3
= -3(1 – cos2θ) + 7 sin2θ
= -3 sin2θ + 7 sin2θ
= 4 sin2θ
Lower limit: x = 3 ⇒ θ = 0
Upper limit: x = 7 ⇒ θ = \(\frac{\pi}{2}\)
Question 11.
Show that the area enclosed between the curves y2 = 12(x + 3) and y2 = 20(5 – x) is \(64 \sqrt{\frac{5}{3}}\).
Solution:
Given curves are y2 = 12(x + 3)
y = \(2 \sqrt{3} \sqrt{x+3}\) ……(1)
y2 = 20(5 – x)
y = \(2 \sqrt{5} \sqrt{5-x}\) ……..(2)
Solving (1) and (2)
\(2 \sqrt{3} \sqrt{x+3}\) = \(2 \sqrt{5} \sqrt{5-x}\)
squaring on both sides
12(x + 3) = 20(5 – x)
⇒ 3x + 9 = 25 – 5x
⇒ 8x = 16
⇒ x = 2
Question 12.
Show that the area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) (ellipse) is πab. Also, deduce the area of the circle x2 + y2 = a2. [(AP) May ’17; Mar. ’14]
Solution:
Given ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
∴ Required area = πab sq. units
If b = a, the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) becomes circle as, x2 + y2 = a2.
Area of circle = πa(a) = πa2 (since b = a).
Question 13.
Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with OA = a, OB = b. Then show that the area bounded between the chord AB and the arc AB of the ellipse is \(\frac{(\pi-2) a b}{4}\).
Solution:
Given equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Since A = (a, 0) and B = (0, b).
We have the equation of chord AB is \(\frac{x}{a}+\frac{y}{b}=1\)
Question 14.
m Find the area bounded between the curves y2 = 4ax, x2 = 4by. [(AP) May ’19]
Solution:
Given curves are
y2 = 4ax ………(1)
⇒ y = \(\sqrt{a} \sqrt{x} 2\)
x2 = 4by ………(2)
⇒ y = \(\frac{x^2}{4 b}\)
Question 15.
Find the area bounded between the curves y2 = 4ax, x2 = 4ay.
Solution:
Given curves are y2 = 4ax
⇒ y = \(2 \sqrt{a} \sqrt{x}\) …….(1)
x2 = 4ay
⇒ y = \(\frac{x^2}{4 a}\) ………(2)
Solving (1) & (2)
\(2 \sqrt{a} \sqrt{x}\) = \(\frac{x^2}{4 a}\)
squaring on both sides
4ax = \(\frac{x^4}{16 a^2}\)
⇒ x4 = 64a3x
⇒ x4 – 64a3x = 0
⇒ x(x3 – 64a3) = 0
⇒ x = 0 (or) x3 – 64a3 = 0
⇒ x3 = 64a3
⇒ x = 4a
∴ x = 0, 4a
Question 16.
Evaluate \(\int_a^b \sqrt{(\mathbf{x}-\mathbf{a})(\mathbf{b}-\mathbf{x})} \mathbf{d x}\). [Mar. ’18 (TS)]
Solution:
Put x = a cos2θ + b sin2θ
dx = [a(2 cos θ) (-sin θ) + b 2 sin θ cos θ] dθ
= [-a sin 2θ + b sin 2θ] dθ
= (b – a) sin 2θ dθ
Now, x – a = a cos2θ + b sin2θ – a
= -a(1 – cos2θ) + b sin2θ
= -a sin2θ + b sin2θ
= (b – a) sin2θ
b – x = b – a cos2θ – b sin2θ
= b(1 – sin2θ) – a cos2θ
= b cos2θ – a cos2θ
= (b – a) cos2θ
Lower limit: x = a ⇒ θ = 0
Upper limit: x = b ⇒ θ = π/2
Question 17.
Find \(\int_{-a}^a\left(x^2+\sqrt{a^2-x^2}\right) d x\)
Solution:
Question 18.
Evaluate \(\int_0^\pi x \sin ^3 x d x\)
Solution:
Question 19.
Find \(\int_0^\pi x \sin ^7 x \cos ^6 x d x\). [Mar. ’19 (TS)]
Solution:
Question 20.
If In = \(\int_0^{\pi / 2} \sin ^n x d x\), then show that In = \(\frac{n-1}{n} I_{n-2}\). [Mar. ’15 (TS)]
Solution:
Question 21.
If In = \(\int_0^{\pi / 2} \cos ^n x d x\), then show that In = \(\frac{\mathbf{n}-1}{n} I_{n-2}\)
Solution:
Question 22.
The circle x2 + y2 = 8 is divided into two parts by the parabola 2y = x2. Find the area of both parts.
Solution:
Given, the circle x2 + y2 = 8 ……..(1)
⇒ y = \(\sqrt{8-x^2}\)
The parabola is 2y = x2 ……..(2)
⇒ y = \(\frac{x^2}{2}\)
Solving (1) and (2)
2y + y2 = 8
⇒ y2 + 2y – 8 = 0
⇒ (y + 4)(y – 2) = 0
⇒ y = -4 or y = 2
From (1) ⇒ x2 + (-4)2 = 8
x2 = 8 – 16
-8 ≠ 8
From (2) ⇒ x2 = 2(2)
∴ x = ±2
Question 23.
If In = \(\int_0^{\pi / 4} \tan ^n x d x\), then show that In+ In-2 = \(\frac{1}{n-1}\). [Mar. ’06, ’99, ’98]
Solution:
Put tan x = t
sec2x dx = dt
Lower limit: x = 0 ⇒ t = 0
Upper limit: x = \(\frac{\pi}{4}\) ⇒ t = 1
∴ In + In-2 = \(\frac{1}{n-1}\)
Question 24.
If \(I_{m, n}=\int_0^{\pi / 2} \sin ^m x \cos ^n x d x\), then show that \(I_{m, n}=\frac{m-1}{m+n} I_{m-2, n}\). [May ’99]
Solution:
Question 25.
Evaluate \(\int_2^6 \sqrt{(6-x)(x-2)} d x\)
Solution:
Put x = 2cos2θ + 6sin2θ then
dx = [2(2) cos θ . (-sin θ) + 6(2) sin θ cos θ] dθ
= [-(2) . 2 sin θ cos θ + 6(2) sin θ cos θ] dθ
= [-2 . sin 2θ + 6 sin 2θ] dθ
= 4 sin 2θ dθ
6 – x = 6 – 2 cos2θ – 6 sin2θ
= 6(1 – sin2θ) – 2 cos2θ
= 6 cos2θ – 2 cos2θ
= 4 cos2θ
x – 2 = 2 cos2θ + 6 sin2θ – 2
= -2(1 – cos2θ) + 6 sin2θ
= -2 sin2θ + 6 sin2θ
= 4 sin2θ
Lower limit: x = 2 ⇒ θ = 0
Upper limit: x = 6 ⇒ θ = \(\frac{\pi}{2}\)
Question 26.
Evaluate \(\int_4^9 \frac{d x}{\sqrt{(9-x)(x-4)}}\)
Solution:
Put x = 4 cos2θ + 9 sin2θ then
dx = [4(2) cos θ (-sin θ) + 9(2) sin θ . cos θ] dθ = 0
= (-4 sin 2θ + 9 sin 2θ) dθ
= 5 sin 2θ dθ
9 – x = 9 – 4 cos2θ – 9 sin2θ
= 9(1 – sin2θ) – 4 cos2θ
= 9 cos2θ – 4 cos2θ
= 5 cos2θ
x – 4 = 4 cos2θ + 9 sin2θ – 4
= -4 (1 – cos2θ) + 9 sin2θ
= -4 sin2θ + 9 sin2θ
= 5 sin2θ
Lower limit: x = 4 ⇒ θ = 0
Upper limit: x = 9 ⇒ θ = \(\frac{\pi}{2}\)