TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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 \(\sec ^2 \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q1

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
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q2
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q2.1

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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 = \(\frac{\pi}{4}\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q3
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q3.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q3.2

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:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q4
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q4.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q4.2
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q4.3

Question 5.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^2 x}{\cos x+\sin x} d x\). [(TS) May ’15]
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q5
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q5.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q5.2
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q5.3

Question 6.
Evaluate \(\int_0^\pi \frac{x}{1+\sin x} d x\)
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q6
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q6.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q6.2

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q7
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q7.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q7.2

Question 8.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\). [(AP) May ’18, ’16, ’14]
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q8
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q8.1

Question 9.
Evaluate \(\frac{x \sin ^3 x}{1+\cos ^2 x} d x\). [(TS) May ’18; Mar. ’15]
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q9
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q9.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q9.2

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 = \(\frac{\pi}{2}\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q10

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q11
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q11.1

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 \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q12
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q12.1
∴ 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 \(\frac{x}{a}+\frac{y}{b}=1\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q13
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q13.1

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 = \(\frac{x^2}{4 b}\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q14
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q14.1

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)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q15
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
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L1 Q15.1

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q1.1

Question 17.
Find \(\int_{-a}^a\left(x^2+\sqrt{a^2-x^2}\right) d x\)
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q2

Question 18.
Evaluate \(\int_0^\pi x \sin ^3 x d x\)
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q3
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q3.1

Question 19.
Find \(\int_0^\pi x \sin ^7 x \cos ^6 x d x\). [Mar. ’19 (TS)]
Solution:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q4
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q4.1

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:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q5

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q6
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q6.1

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
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q7
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q7.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q7.2

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:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q8
Put tan x = t
sec2x dx = dt
Lower limit: x = 0 ⇒ t = 0
Upper limit: x = \(\frac{\pi}{4}\) ⇒ t = 1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q8.1
∴ In + In-2 = \(\frac{1}{n-1}\)

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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:
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q9
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q9.1
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q9.2

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 = \(\frac{\pi}{2}\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q10
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q10.1

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

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 = \(\frac{\pi}{2}\)
TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type L2 Q11

Leave a Comment