Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.5

Integrate the following functions with respect to x

Question 1.

Question 2.

Question 3.
(2x – 5) (3x + 4x)
∫(2x – 5) (3x + 4x) dx
= ∫(72 x + 8x2 – 180 – 20x) dx
= ∫ 72 x dx + ∫82x2 dx – ∫180 dx – ∫2o x dx
= 72∫x dx + 8∫x2dx – 180 ∫dx – 20 ∫x dx

Question 4.
cot2 x + tan2 x
∫cot2 x + tan2 x
= ∫(cosec2x – 1 + sec2x – 1) dx
= ∫(cosec2x + sec2x – 2) dx
= ∫cosec2x dx + ∫sec2x dx – ∫2 dx
= – cot x + tan x – 2x + c
= tan x – cot x – 2x + c

Question 5.

[cos 2x = cos2x – sin2x-1
cos 2x = 2 cos2x – 1]

= 2 ∫(cos x + cos α) dx
= 2 ∫ cos x dx + 2 ∫ cos α dx
= 2 ∫ cos x dx + 2 ∫ cos α ∫ dx
= 2 sin x + 2 cos α(x) + c
= 2 sin x + 2x cos α + c

Question 6.

Question 7.
$$\frac{3+4 \cos x}{\sin ^{2} x}$$

= 3 ∫ cosec2 x dx + 4 ∫ cot x cosec x . dx
= 3 ∫ cosec2 x dx + 4 ∫ cosec x cot x dx
= 3 × – cot x + 4 × – cosec x + c
= – 3 cot x – 4 cosec x + c

Question 8.
$$\frac{\sin ^{2} x}{1+\cos x}$$

= ∫ (1 – cos x) dx
= ∫ dx – ∫ cos x dx
= x – sin x + cannot

Question 9.
$$\frac{\sin 4 x}{\sin x}$$

[sin 2A = 2 sin A cos A]

= 4 ∫ cos 2x cos x . dx
= 2 ∫ 2 cos 2x cosx . dx
= 2 ∫ [cos(2x + x) + cos(2x – x)] dx
[2 cos A cos B = cos (A + B) + cos (A – B)]
= 2 ∫ (cos 3x + cos x) dx
= 2 ∫ cos 3x dx + 2 ∫ cos x dx
= 2 $$\frac{\sin 3 x}{3}$$ + 2 sin x + c
= 2 [$$\frac{\sin 3 x}{3}$$ + sin x] + c

Question 10.
cos 3x cos 2x

Question 11.
sin2 5x

Question 12.

[sin 2A = 2 sin A cos A]

Question 13.
ex log a ex
∫ex log a ex = ∫e log ax . ex . dx
= ∫ ax ex . dx
= ∫ (ae)x . dx
[∫ax . dx = $$\frac{\mathrm{a}^{x}}{\log \mathrm{a}}$$ + c]
= $$\frac{(\mathrm{ae})^{x}}{\log (\mathrm{ae})}$$ + c

Question 14.

Question 15.

Question 16.

Question 17.

Put x = – 3
– 3 + 1 = A (- 3 + 3) + B (- 3 + 2)
– 2 = A × 0 + B (- 1)
B = 2

Put x = – 2
– 2 + 1 = A(- 2 + 3) + B(- 2 + 2)
– 1 = A × 1 + B × 0
A = – 1

= – log |x + 2| + 2 log |x + 3| + c
= 2 log |x + 3| – log |x + 2|+ c

Question 18.

1 = A(x + 2)2 + B (x – 1) (x + 2) + C (x – 1)

Put x= -2
1 = A(- 2 + 2)2 + B(- 2 – 1) (- 2 + 2) + C(- 2 – 1)
1 = A × 0 + B × 0 + C × – 3
C = $$\frac{1}{3}$$

Put x = 1
1 = A(1 + 2)2 + B (1 – 1) (1 + 2) + C(1 – 1)
1 = A × 32 + B × 0 + C × 0
A = $$\frac{1}{9}$$

Put x = 0
1 = A (0 + 2)2 + B (0 – 1) (0 + 2) + C (0 – 1)
1 = A × 4 – 2B – C
1 = $$\frac{1}{9}$$ × 4 – 2B + $$\frac{1}{3}$$
1 = $$\frac{4}{9}+\frac{1}{3}$$ – 2B

Question 19.

3x – 9 = A(x + 2) (x2 + 1) + B(x – 1) (x2 + 1) + (Cx + D) (x – 1) (x + 2)
3x – 9 = A (x + 2) (x2 + 1) + B(x – 1) + (x2 + 1) + Cx (x – 1) (x + 2) + D (x – 1) (x + 2)

Put x = – 2
3 × – 2 – 9 = A (- 2 + 2) ((2)2 + 1) + B(- 2 – 1) ((- 2)2 + 1) + C(- 2) (- 2 – 1) (- 2 + 2) + D(- 2 – 1) (- 2 + 2)
– 6 – 9 = A × 0 + B × (-3) (4 + 1) + C × 0 + D × 0
-15 = B ×- 3 × 5
-15 = – 15B ⇒ B = 1

Put x = 1
3 × 1 – 9 = A(1 + 2) (12 + 1) + B(1 – 1) (12 + 1) + C × 1 (1 – 1) (1 + 2) + D(1 – 1) (1 + 2)
3 – 9 = A × 3 × 2 + B × 0 + C × 0 + D × 0
– 6 = 6A ⇒ A = – 1

Put x = 0
3 × 0 – 9 = A(0 + 2) (02 + 1) + B(0 – 1) (02 + 1) + C × 0 (0 – 1) (0 + 2) + D(0 – 1) (0 + 2)
– 9 = 2A – B + 0 – 2D
– 9 = 2A – B – 2D
– 9 = – 2 × – 1 – 1 – 2D
– 9 = – 2 – 1 – 2D
9 = 3 + 2D
⇒ 2D = 9 – 3
⇒ 2D = 6 ⇒ D = 3

Put x = – 1
3 × – 1 – 9 = A(- 1 + 2) ((1)2 + 1) + B(- 1 – 1) ((- 1)2 + 1)) + C × – 1 × (- 1 – 1) (- 1 + 2) + D(- 1 – 1) (- 1 + 2)
– 3 – 9 = A × 1(1 + 1)+ B × (- 2) (1 + 1) – C × – 2 + D × – 2 × 1
– 12 = 2A – 4B + 2C – 2D
– 12 = 2 × -1 – 4 × 1 + 2C – 2 × 3
– 12 = – 2 – 4 + 2C – 6
– 12 = – 12 + 2C ⇒ C = 0

Question 20.