Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals

Plus Two Maths Integrals Three Mark Questions and Answers

Question 1.
Integrate the following. (3 Score each)

1. ∫sin x sin 2x sin 3 xdx
2. ∫sec2x cos22x dx

Answer:
1. We have sinxsin2xsin3x
= 1/2 (2sinxsin3x) sin2x
= 1/2 (cos2x – cos4x) sin2x
= 1/4 (2sin2xcos2x – 2cos4xsi n2x)
= 1/4 [sin4x – (sin6x – sin2x)]
= 1/4(sin4x + sin2x – sin6x)
∫sin x sin 2x sin 3 xdx
= \(\frac{1}{4}\) ∫(sin 4x + sin 2x – sin 6x)dx
= –\(\frac{1}{16}\) cos4x – \(\frac{1}{8}\) cos2x + \(\frac{1}{24}\) cos6x + c.

2. sec2x cos22x = \(\frac{\left(2 \cos ^{2} x-1\right)^{2}}{\cos ^{2} x}\)
= \(\left(\frac{2 \cos ^{2} x}{\cos x}-\frac{1}{\cos x}\right)^{2}\) = (2cosx – secx)2
= 4cos2x + sec2x – 4
= 2(1 + cos2x) + sec2x – 4
= 2cos2x + sec2x – 2
∫sec2 x cos2 2x dx = ∫(2 cos 2x + sec2 x – 2)dx
= sin 2x + tan x – 2x + c.

 

 

  

 

 

Question 5.
(i) If f (x) is an odd function, then \(\int_{-a}^{a} f(x)\) = ?
(a) 0
(b) 1
(c) 2\(\int_{0}^{a} f(x)\) dx
(d) 2a
Evaluate
(ii) \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x\)
(iii) \(\int_{-1}^{1} e^{|x|} d x\)
Answer:
(i) (a) 0.

(ii) Here, f(x) = sin99x.cos100x .then,
f(-x) = sin99(- x).cos100(- x) = – sin99 x. cos100 x = -f(x)
∴ odd function ⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x=0\).

(iii) Here, f(x) = e|x|, f(-x) = e|-x| = e|x| = f(x)
∴ even function.

we have |x| = x, 0 ≤ x ≤ 1

Question 6.

1. Show that cos2 x is an even function. (1)
2. Evaluate \(\int_{-\pi / 4}^{\pi / 4} \cos ^{2} x d x\) (2)

Answer:
1. Let f(x) = cos2x ⇒ f(-x) = cos2 (-x) = cos2 x = f(x) even.

2.

 

Question 8.
Find the following integrals.
    

 

Question 9.
Find the following integrals.

1. \(\int \frac{1}{3+\cos x} d x\)
2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\)

Answer:
1. \(\int \frac{1}{3+\cos x} d x\)
Put t = tanx/2 ⇒ dt = 1/2 sec2 x/2 dx

2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\) = \(\int \frac{2 x}{(x+2)(x+1)} d x\)

2x = A(x + 1) + B (x + 2)
when x = -1, -2 = B ; B = -2
when x = -2, -4 = -A ; A = 4

= 4log(x + 2) – 2log (x + 1) + C.

Plus Two Maths Integrals Four Mark Questions and Answers

      

Question 4.

1. Choose the correct answer from the bracket.
   ∫ex dx = — (e2x + c, e-x + c, e2x + c) (1)
2. Evaluate: ∫ ex sinxdx

Answer:
1. ex + c

2. I = ∫ex sinxdx = sinx.ex – ∫cos x.exdx
= sin x.ex – (cos x.ex – ∫(- sin x).ex dx)
= sinx.ex – cosxex – ∫sinx.exdx
= sin x.ex – cos xex – I
2I = sin x.ex – cos xex
I = \(\frac{1}{2}\)ex(sinx – cosx) + c.

Question 5.
(i) f(x)∫g(x) dx – ∫(f'(x)∫g(x) dx)dx (1)
(a) ∫f'(x)g{x)dx
(b) ∫f(x)g'(x)dx
(c) ∫\(\frac{f(x)}{g(x)}\)dx
(d) ∫f(x)g(x)dx
(ii) Integrate sin-1\(\sqrt{\frac{x}{a+x}}\)dx w.r.to x. (3)
Answer:
(i) (d) ∫f(x)g(x)dx

(ii) ∫sin-1\(\sqrt{\frac{x}{a+x}}\)dx,
Put x = a tan2θ, θ = tan-1\(\sqrt{\frac{x}{a}}\)
⇒ dx = 2a tanθ sec2θ dθ
I = ∫sin-1\(\left(\frac{\tan \theta}{\sec \theta}\right)\) 2a tanθ sec2θ dθ
= ∫sin-1(sinθ)2a tanθ sec2θ dθ
= 2a∫θ tanθ sec2θ dθ
Put tanθ = t, θ = tan-1 t ⇒ sec2θ dθ = dt
= 2a ∫ tan-1 t (t) dθ

   

2. ∫sec x(sec x + tan x)dx = ∫(sec2 x + sec x. tan x)dx
= tanx + secx + c.

3. ∫e3xdx = \(\frac{e^{3 x}}{3}\) + c.

4. ∫(sin x + cos x)dx = sin x – cosx + c.

Question 9.
Consider the integral I = \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)?

1. What substitution can be given for simplifying the above integral? (1)
2. Express I in terms of the above substitution. (1)
3. Evaluate I. (2)

Answer:
1. Substitute sin-1 x = t.

2. We have, sin-1 x = t ⇒ x = sint
Differentiating w.r.t. x; we get,
\(\frac{1}{\sqrt{1-x^{2}}}\)dx = dt
∴ I = ∫t sin t dt.

3. I = ∫t sin t dt = t.(-cost) -∫(-cost)dt = -t cost + sint + c
= -sin-1 x. cos (sin-1 x) + sin(sin-1 x) + c
x – sin-1 x.cos(sin-1 x) + c.

Question 10.
Evaluate \(\int_{0}^{\pi / 4} \log (\tan x) d x\).
Answer:

Question 11.
Find the following integrals.

1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) (2)
2. \(\int \frac{1}{x^{2}-6 x+13} d x\) (2)

Answer:
1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) = \(\int \frac{\sin ^{2} x}{\cos ^{2} x} d x\) = ∫tan2 xdx
= ∫(sec2x – 1)dx = tanx – x + c.

2. \(\int \frac{1}{x^{2}-6 x+13} d x\)

 

Question 13.
(i) ∫sin2x dx = ? (1)
(a) 2 cos x + c
(b) -2 sin x + c
(c) \(\frac{\cos 2 x}{2}\) + c
(d) \(-\frac{\cos 2 x}{2}\) + c
(ii) Evaluate ∫ex sin 2x dx (3)
Answer:
(i) (d) \(-\frac{\cos 2 x}{2}\) + c.

(ii) Consider I = ∫ex sin 2x dx
= ∫sin 2x. exdx = sinx.ex – 2∫cos 2x. exdx
= sin 2x.ex – 2 (cos 2x.ex + 2∫sin 2x. exdx)
= sin 2x. ex – 2 cos 2x ex – 4 ∫sin 2x. exdx
= sin 2x. ex – 2 cos 2x ex – 4I
5 I = sin 2x. ex – 2 cos 2x ex
I = \(\frac{e^{x}}{5}\) (sin 2x – 2 cos 2x).

     

Plus Two Maths Integrals Six Mark Questions and Answers

Question 1.
(i) Fill in the blanks. (3)
(a) ∫ tan xdx = —
(b) ∫ cos xdx = —
(c) ∫\(\frac{1}{x}\)dx = —
(ii) Evaluate ∫sin3 xcos2 xdx (3)
Answer:
(i) (a) log|secx| + c
(b) sinx + c
(c) log|x| + c.

(ii) ∫sin3 xcos2 xdx = ∫sin2 xcos2 x sin xdx
= ∫(1 – cos2 x)cos2 x sin xdx
Put cos x = t ⇒ – sin xdx = dt
∴ ∫(1 – cos2 x)cos2 xsin xdx = -∫(1 – t2 )t2dt
= ∫(t4 – t2)dt = \(\frac{t^{5}}{5}-\frac{t^{3}}{3}\) + c
= \(\frac{\cos ^{5} x}{5}-\frac{\cos ^{3} x}{3}\) + c.

     

  

Question 5.

1. Evaluate the as \(\int_{0}^{2}\)x2dx the limit of a sum. (3)
2. Hence evaluate \(\int_{-2}^{2}\)x2dx (1)
3. If \(\int_{0}^{2}\) f(x)dx = 5 and \(\int_{-2}^{2}\) f(x)dx = 0, then \(\int_{-2}^{0}\) f(x)dx = …….. (2)

Answer:
1. Here the function is f(x) = x2, a = 0, b = 2 and h = \(\frac{b-a}{n}=\frac{2}{n}\)
\(\int_{0}^{2}\)x2dx =

        

   

Question 11.

1. Find ∫\(\frac{1}{x^{2}+a^{2}}\)dx (1)
2. Show that 3x + 1 = \(\frac{3}{4}\)(4x – 2) + \(\frac{5}{2}\) (2)
3. Evaluate \(\int \frac{3 x+1}{2 x^{2}-2 x+3} d x\) (3)

Answer:
1. ∫\(\frac{1}{x^{2}+a^{2}}\)dx = 1/a tan-1 x/a + c.

2. 3x + 1 = A \(\frac{d}{d x}\)(2x2 – 2x + 3) + B
= A(4x – 2) + B
3 = 4A; A = 3/4
1 = -2A + B
1 = -3/2 + B, B = 1 + 3/2 = 5/2
∴ 3x + 1 = 3/4(4x – 2) + 5/2

3.