Samacheer Kalvi 11th Maths Guide Chapter 11 Integral Calculus Ex 11.5

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.

Answer:

Question 2.

Answer:

Question 3.
(2x – 5) (3x + 4x)
Answer:
∫(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
Answer:
∫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.

Answer:

[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.

Answer:

Question 7.
\(\frac{3+4 \cos x}{\sin ^{2} x}\)
Answer:

= 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.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Answer:

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

Question 9.
\(\frac{\sin 4 x}{\sin x}\)
Answer:

[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
Answer:

Question 11.
sin² 5x
Answer:

Question 12.

Answer:

[sin 2A = 2 sin A cos A]

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

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:


1 = A(x – 2) + B(x – 1)

put x = 2
1 = A(2 – 2) + B(2 – 1)
1 = A × 0 + B × 1
B = 1

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