Samacheer Kalvi 11th Maths Guide Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 கணங்கள், தொடர்புகள் மற்றும் naசார்புகள் Ex 11.2 Textbook Questions and Answers, Notes.

TN Board 11th Maths Solutions Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.2

x -ஐப் பொறுத்து கீழ்க்காண்பவற்றைத் தொகையிடுக.

Question 1.
(i) (x + 5)⁶

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

(iii) \(\sqrt{3 x+2}\)
தீர்வு :
(i) (x + 5)⁶
∫ (x + 5)⁶ dx = \(\frac{(x+5)^{6+1}}{6+1}\) + c

= \(\frac{(x+5)^{7}}{7}\) + c

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

∫ \(\frac{1}{(2-3 x)^{4}}\) dx = ∫ (2 – 3x)^-4 dx

= \(\frac{(2-3 x)^{-4+1}}{(-4+1)(-3)}\)

= \(\frac{(2-3 x)^{-3}}{9}\)

= \(\frac{1}{9(2-3 x)^{3}}\) + c

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

∫ \(\sqrt{3 x+2}\) dx = ∫ (3x + 2)^{\(\frac{1}{2}\)} dx
= \(\frac{(3 x+2)^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)(3)}\) + c

= \(\frac{(3 x+2)^{\frac{3}{2}}}{\frac{3}{2} \times 3}\) + c

= \(\frac{2}{9}\) (3x + 2)^{\(\frac{3}{2}\)} + c

Question 2.
(i) sin 3x
(ii) cos(5 – 11x)
(iii) cosec²(5x – 7)
தீர்வு :

(i) ∫ sin 3x dx = \(-\frac{\cos 3 x}{3}\) + c

(ii) ∫ cos(5 – 11x) dx = sin \(\frac{(5-11 x)}{-11}\) + c

= \(\frac{-1}{11}\) sin(5 – 11x) + c

(iii) ∫ cosec² (5x – 7) dx = –\(\frac{\cot (5 x-7)}{5}\) + c

Question 3.
(i) e^{3x – 6}
(ii) e^{8 – 7x}
(iii) \(\frac{1}{6-4 x}\)
தீர்வு :

(i) ∫ e^{3x – 6} dx = \(\frac{e^{3 x-6}}{3}\) + c
= \(\frac{1}{3}\) e^{3x – 6} + c

(ii) ∫ e^{8 – 7x} dx

= \(\frac{e^{8-7 x}}{(-7)}\) + c

= –\(\frac{1}{7}\) e^{8 – 7x} + c

(iii) ∫ \(\frac{1}{6-4 x}\) dx

= \(\frac{\log |6-4 x|}{-4}\) + c

= –\(\frac{1}{4}\) log |6 – 4x| + c

Question 4.
(i) sec² \(\frac{x}{5}\)

(ii) cosec (5x + 3) cot (5x + 3)

(iii) 30 sec (2 – 15x) tan (2 – 15x)
தீர்வு :

(i) ∫ sec² \(\frac{x}{5}\) dx

= \(\frac{\tan \left(\frac{x}{5}\right)}{\frac{1}{5}}\)

= 5 tan (\(\frac{x}{5}\)) + c

(ii) ∫ cosec (5x + 3) cot (5x + 3) dx = -cosec \(\frac{(5 x+3)}{5}\) + c

(iii) ∫ 30 sec (2 – 15x) tan (2 – 15x) dx
= 30 ∫ sec (2 – 15x) tan (2 – 15x) dx
= 30 \(\frac{\sec (2-15 x)}{-15}\) + cos
= -2 sec(2 – 15x) + c

Question 5.
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)

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

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

(i) ∫ \(\frac{1}{\sqrt{1-(4 x)^{2}}}\) dx = sin^-1 x + c

∫ \(\frac{1}{\sqrt{1-(4 x)^{2}}}\) dx = \(\frac{\sin ^{-1}(4 x)}{4}\) + c
= \(\frac{1}{4}\) sin^-1 (4x) + c

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

= ∫ \(\frac{d x}{\sqrt{1-(9 x)^{2}}}\) = \(\frac{\sin ^{-1}(9 x)}{9}\) + c

= \(\frac{1}{9}\) sin^-1 (9x) + c

(iii) ∫ \(\frac{1}{1+36 x^{2}}\) dx = ∫ \(\frac{1}{1+(6 x)^{2}}\) dx

= \(\frac{\tan ^{-1}(6 x)}{6}\)

= \(\frac{1}{6}\) tan^-1 (6x) + c

[∵ ∫ \(\frac{d x}{1+x^{2}}\) = tan^-1 + c]