Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.2

Integrate the following functions with respect to x

Question 1.
(i) (x + 5)6
(ii) $$\frac{1}{(2-3 x)^{4}}$$
(iii) $$\sqrt{3 x+2}$$
Answer:
(i) (x + 5)6

(ii) $$\frac{1}{(2-3 x)^{4}}$$

(iii) $$\sqrt{3 x+2}$$

Question 2.
(i) sin 3x
(ii) cos (5 – 11x)
(iii) cosec2 (5x – 7)
Answer:
(i) sin 3x
∫sin (ax + b) dx = – $$\frac{1}{a}$$ cos (ax + b) + c
∫sin 3x dx = – $$\frac{1}{3}$$ cos 3x + c

(ii) cos (5 – 11x)

(iii) cosec2 (5x – 7)
[∫cosec2 (ax + b) dx = – $$\frac{1}{a}$$ cot (ax + b) + c]
∫cosec2 (5x – 7) dx = – $$\frac{1}{5}$$ cot (5x – 7) + c

Question 3.
(i) e3x – 6
(ii) e8 – 7x
(iii) $$\frac{1}{6-4 x}$$
Answer:
(i) e3x – 6
[∫eax+b dx = $$\frac{1}{a}$$ eax + b + c]
∫e3x – 6 dx = $$\frac{1}{3}$$ e3x – 6 + c

(ii) e8 – 7x
[∫eax+b dx = $$\frac{1}{a}$$ eax + b + c]
∫e8 – 7x = $$\frac{1}{-7}$$ e8 – 7x + c
∫e8 – 7x = $$-\frac{1}{7}$$ e8 – 7x + c

(iii) $$\frac{1}{6-4 x}$$

Question 4.
(i) sec2 $$\frac{x}{5}$$
(ii) cosec (5x + 3) cot (5x + 3)
(iii) 30 sec (2 – 15x) tan (2 – 15x)
Answer:
(i) sec2 $$\frac{x}{5}$$
[∫sec2 (ax + b)dx = $$\frac{1}{a}$$ tan (ax + b) + c]

(ii) cosec (5x + 3) cot (5x + 3)
[∫cosec (ax + b) cot (ax + b) dx = – $$\frac{1}{a}$$ cosec (ax + b) + c]
∫cosec (5x + 3) cot (5x + 3) dx = – $$\frac{1}{5}$$ cosec (5x + 3) + c]

(iii) 30 sec (2 – 15x) tan (2 – 15x)
[∫sec (ax + b) tan (ax + b)dx = $$\frac{1}{a}$$ sec (ax + b) + c]
∫30 sec (2 – 15x) tan (2 – 15x)dx = 30 × $$\frac{1}{-15}$$ × sec (2 – 15x) + c
= – 2 sec (2 – 15x) + c

Question 5.
(i) $$\frac{1}{\sqrt{1-(4 x)^{2}}}$$
(ii) $$\frac{1}{\sqrt{1-81 x^{2}}}$$
(iii) $$\frac{1}{1+36 x^{2}}$$
Answer:
(i) $$\frac{1}{\sqrt{1-(4 x)^{2}}}$$

(ii) $$\frac{1}{\sqrt{1-81 x^{2}}}$$

(iii) $$\frac{1}{1+36 x^{2}}$$