Samacheer Kalvi 11th Maths Guide Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.6

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 கணங்கள், தொடர்புகள் மற்றும் naசார்புகள் Ex 11.6 Textbook Questions and Answers, Notes.

TN Board 11th Maths Solutions Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.6

கீழ்க்காண்பவற்றைத் தொகையிடுக.

Question 1.
\(\frac{x}{\sqrt{1+x^{2}}}\)
தீர்வு :
I = ∫ \(\frac{x}{\sqrt{1+x^{2}}}\) dx என்க

u = 1 + x²
⇒ du = 2x dx ⇒ \(\frac{d u}{2}\) = x dx
∴ I = ∫ \(\frac{\frac{d u}{2}}{\sqrt{u}}\) dxz

= \(\frac{1}{2} \int \frac{d u}{\sqrt{u}}\)

= \(\frac{1}{2}\) ∫ u^{\(-\frac{1}{2}\)} du

= \(\frac{1}{2} \frac{u^{-\frac{1}{2}+1}}{\left(\frac{-1}{2}+1\right)}\) + c

=

= √u + c = \(\sqrt{1+x^{2}}\) + c [∵ u = 1 + x²]

Question 2.
\(\frac{x^{2}}{1+x^{6}}\)
தீர்வு :
I = ∫ \(\frac{x^{2}}{1+x^{6}}\) dx என்க

∴ I = ∫ \(\frac{x^{2} d x}{1+\left(x^{3}\right)^{2}}\)

u = x³ ⇒ du = 3x² dx
⇒ \(\frac{d u}{3}\) = x² dx

∴ I = ∫ \(\frac{\frac{d u}{3}}{1+u^{2}}\)

= \(\frac{1}{3}\) ∫ \(\frac{d u}{1+u^{2}}\)

= \(\frac{1}{3}\) tan^-1 (u) + c

= \(\frac{1}{3}\) tan^-1 (x³) + c [∵ u = x³]

Question 3.
\(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)
தீர்வு :
I = \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\) dx என்க
u = e^x + e^-x
⇒ du = (e^x – e^-x) dx
∴ I = ∫ \(\frac{d u}{u}\) = log |u| + c
= log |e^x e^-x| + c

Question 4.
\(\frac{10 x^{9}+10^{x} \log _{e} 10}{10^{x}+x^{10}}\)
தீர்வு :
I = ∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{10^{x}+x^{10}}\) dx

∴ u = 10^x + x¹⁰

⇒ du = (10x⁹ + 10^x . log_e 10)dx

∴ I = ∫ \(\frac{d u}{u}\) = log |u| = log|10^x + x¹⁰| + c

Question 5.
\(\frac{\sin \sqrt{x}}{\sqrt{x}}\)
தீர்வு :
I = ∫ \(\frac{\sin \sqrt{x}}{\sqrt{x}}\) dx என்க

√x = u ⇒ \(\frac{1}{2 \sqrt{x}}\) dx = du

⇒ \(\frac{d x}{\sqrt{x}}\) = 2du

∴ I = 2 ∫ sin u .du = 2 ∫ sin u du
= – 2 cos u + c = -2 cos √x + c

Question 6.
\(\frac{\cot x}{\log (\sin x)}\)
தீர்வு :
I = ∫ \(\frac{\cot x}{\log (\sin x)}\) dx என்க

u = log(sin x) ⇒ du = \(\frac{1}{\sin x}\) cos x dx
⇒ du = cot x dx
∴ I = ∫ \(\frac{d u}{u}\) = log |u| + c = log log(sin x) + c

Question 7.

தீர்வு :

Question 8.
\(\frac{\sin 2 x}{a^{2}+b^{2} \sin ^{2} x}\)
தீர்வு :
I = ∫ \(\frac{\sin 2 x}{a^{2}+b^{2} \sin ^{2} x}\) என்க
u = a² + b² sin² x
⇒ du = b² 2 sin x . cos x dx
⇒ du = b² sin 2x dx
⇒ \(\frac{d u}{b^{2}}\) = sin 2x dx
I = \(\frac{1}{b^{2}} \int \frac{d u}{u}\)
∴ I = \(\frac{1}{b^{2}}\) log |u| + c
⇒ I = \(\frac{1}{b^{2}}\) log |a² + b² sin² x| + c

Question 9.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
தீர்வு :
I = ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx என்க
u = sin^-1 x
⇒ du = \(\frac{1}{\sqrt{1-x^{2}}}\) dx
∴ I = ∫ u . du = \(\frac{u^{2}}{2}\) + c
= \(\frac{\left(\sin ^{-1} x\right)^{2}}{2}\) + c

Question 10.
\(\frac{\sqrt{x}}{1+\sqrt{x}}\)
தீர்வு :
I = ∫ \(\frac{\sqrt{x}}{1+\sqrt{x}}\) dx என்க
1 + √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt

⇒ dx = 2√x dt பதும் √x = t – 1

I = \(\int \frac{(t-1) \cdot 2(t-1) d t}{t}\)

= 2 \(\int \frac{(t-1)^{2}}{t}\) dt

= 2 ∫ \(\frac{t^{2}+1-2 t}{t}\) dt

∴ I = 2 ∫ \(\left(\frac{t^{2}}{t}+\frac{1}{t}-\frac{2 t}{1}\right)\) dt

= 2 ∫ (t + \(\frac{1}{t}\) – 2) dt

= 2 [\(\frac{t^{2}}{2}\) + log |t| – 2t] + c

= t² + 2 log|t| – 4t + c

= (1 + √x)² + 2 log|1 +√x| – 4(1 + √x) + c [∵ t = 1 + √x]

Question 11.
\(\frac{1}{x \log x \log (\log x)}\)
தீர்வு :
I = ∫ \(\frac{1}{x \log x \log (\log x)}\) dx என்க

u = log(log x)

du = \(\frac{1}{\log x} \cdot \frac{1}{x}\) dx

∴ I = ∫ \(\frac{d u}{u}\) = log |u| + c
= log|log(log x)| + c

Question 12.
β x^{α – 1} e^-βxα
தீர்வு :
I = ∫ αβ x^{α – 1} e^-βxα dx என்க
βx^α = u என்க
⇒ β α x^{α – 1} dx = du

∴ I = ∫ e^-u . du = \(\frac{e^{-u}}{-1}\) + c
= e^-u + c
= -e^-βxα + c [∵ u = βx^α]

Question 13.
tan x\(\sqrt{\sec x}\)
தீர்வு :
I = ∫ tan x\(\sqrt{\sec x}\) dx என்க
sec x = u ⇒ sec x tan x dx = du
⇒ tan x dx = \(\frac{d u}{\sec x}\) = \(\frac{d u}{u}\)

∴ I = ∫ √u \(\frac{d u}{u}\) = ∫ \(\frac{d u}{\sqrt{u}}\)

= ∫ u^{\(-\frac{1}{2}\)}

= \(\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + c

= \(\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\) + c = 2√u + c = 2\(\sqrt{\sec x}\) + c [∵ u = sec x]

Question 14.
x(1 – x)¹⁷
தீர்வு :
I = ∫ x(1 – x)¹⁷ dx என்க
u = 1 – x ⇒ x = 1 – u
du = – dx
∴ I = ∫ (1 – u)(u¹⁷) (-du)
= ∫ u¹⁷ – u¹⁸ du
= ∫ u^{18 du – ∫ u17 du}

= \(\frac{u^{19}}{19}-\frac{u^{18}}{18}\) + c

= \(\frac{(1-x)^{19}}{19}-\frac{(1-x)^{18}}{18}\) + c [∵ u = 1 – x]

Question 15.
sin⁵ x cos³ x
தீர்வு :
I = ∫ sin⁵ x cos³ x என்க
= ∫ sin⁵ x . cos ² x . cos x dx
= ∫ sin ⁵ x(1 – sin²) . cos x dx [∵ sin² x + cos ² x = 1]
= ∫ (sin ⁵ x – sin ⁷ x) cos x dx
u = sin x ⇒ du = cos dx
I = ∫ (u⁵ – u⁷) du = ∫ u⁵ dx – ∫ u⁷ dx

= \(\frac{u^{6}}{6}-\frac{u^{8}}{8}\) + c

= \(\frac{\sin ^{6} x}{6}-\frac{\sin ^{8} x}{8}\) + c

Question 16.
\(\frac{\cos x}{\cos (x-a)}\)
தீர்வு :
I = ∫ \(\frac{\cos x}{\cos (x-a)}\) dx

= ∫ \(\frac{\cos (x-a+a)}{\cos (x-a)}\) dx

= ∫ \(\frac{\cos (x-a) \cos a}{\cos (x-a)}\) dx – ∫ \(\frac{\sin (x-a) \sin a}{\cos (x-a)}\) dx

= cos a ∫ dx – sin a ∫ tan(x – a) dx
= cos a(x) – sin a log |sex(x – a)| + c