Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 கணங்கள், தொடர்புகள் மற்றும் naசார்புகள் Ex 11.13 Textbook Questions and Answers, Notes.
TN Board 11th Maths Solutions Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.13
சரியான அல்லது மிகவும் ஏற்புடைய விடையினைக் கொடுக்கப்பட்ட நான்கு மாற்று விடைகளில் இருந்து தேர்ந்தெடுக்கவும்.
Question 1.
∫ f(x)dx = g(x)+ c எனில், ∫ f(x)g'(x) dx என்பது
(1) ∫ (f(x))2
(2) ∫ f(x)g(x) dx
(3) ∫ f'(x)g(x) dx
(4) ∫ (g((x))2
குறிப்பு :
∫ f(x) dx = g(x) + c
f(x) = g'(x)
∫ f(x)g'(x) dx =∫ f(x) × f(x) dx = ∫ [f(x)]2 dx
விடை :
(1) (f(x))
Question 2.
\(\int \frac{3^{\frac{1}{x}}}{x^{2}}\) dx = k(3\(\frac{1}{x}\)) + c எனில், k-ன் மதிப்பு
(1) log 3
(2) -log 3
(3) \(-\frac{1}{\log 3}\)
(4) \(\frac{1}{\log 3}\)
குறிப்பு :
\(\int \frac{3^{\frac{1}{x}}}{x^{2}}\) dx = k (3\(\frac{1}{x}\)) + c
t = 3\(\frac{1}{x}\)
⇒
∫ dt log 3 = k . 3\(\frac{1}{x}\) ⇒ -log 3 . t = k . 3\(\frac{1}{x}\)
– log 3 . 3\(\frac{1}{x}\) = k . 3\(\frac{1}{x}\)
⇒ k = \(-\frac{1}{\log 3}\)
விடை :
(3) \(-\frac{1}{\log 3}\)
Question 3.
f'(x) ex2 dx = (x – 1) ex2 + c எனில், f (x) என்பது
(1) 2x3 – \(\frac{x^{2}}{2}\) + x + c
(2) \(\frac{x^{3}}{2}\) + 3x2 + 4x + c
(3) x3 + 4x2 + 6x + c
(4) 2\(\frac{x^{3}}{3}\) – x2 + x + c
குறிப்பு :
∫ f'(x) ex2 dx = (x – 1) ex2 + c
\(\frac{d}{d x}\) [∫ f'(x) ex2 dx] = \(\frac{d}{d x}\) [(x – 1) (ex2)]
= (x – 1) \(\frac{d}{d x}\) (ex2) + ex2 . \(\frac{d}{d x}\) (x – 1)
f'(x) . ex2 = (x – 1) ex2 (2x) + ex2 (1)
⇒ f'(x) ex2 = ex2 (2x2 – 2x + 1)
⇒ f'(x) = 2x2 – 2x + 1
⇒ ∫ f'(x) dx = ∫ (2x2 – 2x + 1) dx
⇒ f(x) = 2\(\frac{x^{3}}{3}\) – 2\(\frac{x^{2}}{2}\) + x + c
⇒ f(x) = 2\(\frac{x^{3}}{3}\) – x2 + x + c
விடை :
(4) 2\(\frac{x^{3}}{3}\) – x2 + x + c
Question 4.
(x,y) என்ற ஏதேனும் ஒருபுள்ளியில் ஒருவளைவரையின் \(\frac{x^{2}-4}{x^{2}}\) சாய்வு 2 ஆகும். இவ்வளைவரை (2, 7) என்ற புள்ளி வழியாகச் சென்றால், வளைவரையின் சமன்பாடு
(1) y = x + \(\frac{4}{x}\) + 3
(2) y = x + \(\frac{4}{x}\) + 4
(3) y = x2 + 3x + 4
(4) y = x2 -3x + 6
குறிப்பு :
சாய்வு = \(\frac{d y}{d x}=\frac{x^{2}-4}{x^{2}}\)
⇒ f'(x) = \(\frac{x^{2}-4}{x^{2}}\)
= \(\frac{x^{2}}{x^{2}}-\frac{4}{x^{2}}\)
= 1 – \(\frac{4}{x^{2}}\)
= 1 – 4x-2
இருபுறமும் x ஐப் பொறுத்து தொகையீடு காண
∫ f'(x) = ∫ (1 – 4x-2) dx
= x – 4\(\frac{x^{-2+1}}{-2+1}\) + c
f(x) = x – \(\frac{x^{-1}}{-1}\) + c
= x + \(\frac{4}{x}\) + c ………..(1)
வளைவரை (2, 7) வழிச் செல்வதால்
7 = 2 + \(\frac{4}{2}\) + c
⇒ 7 = 2 + 2 + c
⇒ 7 – 4 = c
⇒ c = 3
f(x) = y = x + \(\frac{4}{x}\) + 3 [(1)லிருந்து]
விடை :
(1) y = x + \(\frac{4}{x}\) + 3
Question 5.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)}\) dx =
(1) cot (x ex) + c
(2) sec (x ex) + c
(3) tan (x ex) + c
(4) cos (x ex) + c
குறிப்பு :
I = ∫ ex \(\frac{(1+x)}{\cos ^{2}\left(x e^{x}\right)}\) dx
x ex ⇒ t = (x . ex + ex . 1) d= dt = (x+1) dx = dt
ex (x + 1) dx = dt
∴ I = ∫ \(\frac{d t}{\cos ^{2} t}\)
= ∫ sec2 t dt = tan t + c = tan (x ex) + c
விடை :
(3) tan (x ex) + c
Question 6.
∫ \(\frac{\sqrt{\tan x}}{\sin 2 x}\) dx =
(1) \(\sqrt{\tan x}\) + c
(2) 2\(\sqrt{\tan x}\) + c
(3) \(\frac{1}{2}\) \(\sqrt{\tan x}\) + c
(4) \(\frac{1}{4}\) \(\sqrt{\tan x}\) + c
குறிப்பு :
விடை :
(1) \(\sqrt{\tan x}\) + c
Question 7.
∫ sin3 x dx =
(1) \(-\frac{3}{4}\) cos x – \(\frac{\cos 3 x}{12}\) + c
(2) \(\frac{3}{4}\) cos x + \(\frac{\cos 3 x}{12}\) + c
(3) \(-\frac{3}{4}\) cos x + \(\frac{\cos 3 x}{12}\) + c
(4) \(-\frac{3}{4}\) sin x + \(\frac{\sin 3x}{12}\) + c
குறிப்பு :
I = ∫ sin-3 x dx
sin 3x = 3 sin x – 4 sin3 x
⇒ 4 sin3 x = 3 sin x – sin 3x
⇒ sin3 x = \(\frac{3}{4}\) sin x – \(\frac{1}{4}\) sin 3x
∴ I = \(\frac{3}{4}\) ∫ sin x dx – \(\frac{1}{4}\) ∫ sin 3x dx
= \(\frac{3}{4}\) (-cos x) – \(\frac{1}{4}\) (-\(\frac{\cos 3 x}{3}\)) + c
= –\(\frac{3}{4}\) cos x + \(\frac{\cos 3 x}{12}\) + c
விடை :
(3) \(-\frac{3}{4}\) cos x + \(\frac{\cos 3 x}{12}\) + c
Question 8.
∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx =
(1) x + c
(2) \(\frac{x^{3}}{3}\) + c
(3) \(\frac{3}{x^{3}}\) + c
(4) \(\frac{1}{x^{2}}\) + c
குறிப்பு :
∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx
= ∫ \(\frac{e^{\log x^{6}}-e^{\log x^{5}}}{2^{\log x^{4}}-e^{\log x^{3}}}\) dx
= ∫ \(\frac{x^{6}-x^{5}}{x^{4}-x^{3}}\) dx
= ∫ \(\frac{x^{5}(x-1)}{x^{3}(x-1)}\) dx
= ∫ x2 dx = \(\frac{3}{x^{3}}\) + c
விடை :
(3) \(\frac{3}{x^{3}}\) + c
Question 9.
∫ \(\frac{\sec x}{\sqrt{\cos 2 x}}\) dx =
(1) tan-1 (sin x) + c
(2) 2 sin-1(tan x) + c
(3) tan-1(cos x) + c
(4) sin-1(tan x) + c
குறிப்பு :
I = ∫ \(\frac{d t}{\sqrt{1+t^{2}}}\)
= sin-1 (t) + c = sin-1 (tan x) + c
விடை :
(4) sin-1(tan x) + c
Question 10.
∫ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx =
(1) x2 + c
(2) 2x2 + c
(3) \(\frac{x^{2}}{2}\) + c
(4) – \(\frac{x^{2}}{2}\) + c
குறிப்பு :
∫ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx = ∫ tan-1 \(\left(\sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}}\right)\) dx
[ ∵ cos 2x = 2 cos2 x – 1 மற்றும் cos 2x = 1 – 2 sin2x]
= ∫ tan-1(\(\sqrt{\tan ^{2} x}\)) dx
= ∫ tan-1 (tan x) dx = ∫ x dx = \(\frac{x^{2}}{2}\) + c
விடை :
(3) \(\frac{x^{2}}{2}\) + c
Question 11.
∫ 23x + 5 dx =
(1) \(\frac{3\left(2^{3 x+5}\right)}{\log 2}\) + c
(2) \(\frac{2^{3 x+5}}{2 \log (3 x+5)}\) + c
(3) \(\frac{2^{3 x+5}}{2 \log 3}\) + c
(4) \(\frac{2^{3 x+5}}{3 \log 2}\) + c
குறிப்பு :
∫ 23x + 5 dx = \(\frac{2^{3 x+5}}{\log 2^{3}}\) + c
= \(\frac{2^{3 x+5}}{3 \log 2}\) + c
விடை :
(4) \(\frac{2^{3 x+5}}{3 \log 2}\) + c
Question 12.
∫ \(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\) dx =
(1) \(\frac{1}{2}\) sin 2x + c
(2) –\(\frac{1}{2}\) sin 2x + c
(3) \(\frac{1}{2}\) cos 2x + c
(4) –\(\frac{1}{2}\) cos 2x + c
குறிப்பு :
∫ \(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\) dx = ∫ \(\frac{\left(\sin ^{4} x\right)^{2}-\left(\cos ^{4} x\right)^{2}}{1-2 \sin ^{2} x \cos ^{2} x}\) dx
= ∫ \(\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{4} x-\cos ^{4} x\right)}{1-2 \sin ^{2} x \cos ^{2} x}\) dx
விடை :
(2) –\(\frac{1}{2}\) sin 2x + c
Question 13.
∫ \(\frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1}\) dx =
(1) ex tan-1(x + 1) + c
(2) tan-1 ex + c
(3) ex \(\frac{\left(\tan ^{-1} x\right)^{2}}{2}\) + c
(4) ex tan-1 x + c
குறிப்பு :
∫ \(\frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1}\) dx
= ∫ ex \(\frac{\left(\tan ^{-1} x\left(x^{2}+1\right)+1\right)}{x^{2}+1}\) dx
= ∫ ex \(\left(\frac{\tan ^{-1}(x)\left(x^{2}+1\right)}{x^{2}+1}+\frac{1}{x^{2}+1}\right)\) dx
= ∫ ex \(\left(\tan ^{-1} x+\frac{1}{x^{2}+1}\right)\) dx
f(x) = tan-1 x ⇒ f'(x) = \(\frac{1}{1+x^{2}}\) dx
∴ I ஆன்து ∫ ex [f(x) + f'(x)] dx = ex f(x)
எனும் வடிவில் உள்ளது
I = ex tan-1 x + c
விடை :
(4) ex tan-1 x + c
Question 14.
∫ \(\frac{x^{2}+\cos ^{2} x}{x^{2}+1}\) cosec2 x dx =
(1) cot x + sin-1 x + c
(2) – cot x + tan-1 x + c
(3) – tan x + cot-1 x + c
(4) – cot x – tan-1 x + c
குறிப்பு :
விடை :
(4) – cot x – tan-1 x + c
Question 15.
∫ x2 cos x dx is
(1) x2 sin x + 2x cos x – 2 sin x + c
(2) x2 sin x – 2x cos x – 2 sin x + c
(3) -x2 sin x + 2x cos x + 2 sin x + c
(4) – x2 sin x – 2x cos x + 2 sin x + c
குறிப்பு :
பெர்னோலியின் சூத்திரத்தைப் பயன்படுத்தி
∫ u dv = uv – u’v1 + u”v2 – u”‘v3
u = x2;
u’ = 2x;
u” = 2;
dv = cos x dx;
v = sin x
v1 = – cos x
v2 = – sin x
I = ∫ x2 cos x dx
= x2 sin x – 2x(-cos x) + 2(-sin x) + c
= x2 + 2x cos x – 2 sin x + c
விடை :
(1) x2 sin x + 2x cos x – 2 sin x + c
Question 16.
∫ \(\sqrt{\frac{1-x}{1+x}}\) dx =
(1) \(\sqrt{1-x^{2}}\) + sin-1 x + c
(2) sin-1 x – \(\sqrt{1-x^{2}}\) + c
(3) log |x + \(\sqrt{1-x^{2}}\)| – [/latex] \(\sqrt{1-x^{2}}\) + c
(4) \(\sqrt{1-x^{2}}\) + log|x + \(\sqrt{1-x^{2}}\)| + c
குறிப்பு :
I = ∫ \(\sqrt{\frac{1-x}{1+x}}\) dx
= ∫ \(\sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}}\) dx
= ∫ \(\frac{1-x}{\sqrt{1-x^{2}}}\) dx
= ∫ \(\frac{1}{\sqrt{1-x^{2}}}\) dx – ∫ \(\frac{x}{\sqrt{1-x^{2}}}\) dx
= sin-1 x – ∫ \(\frac{x}{\sqrt{1-x^{2}}}\) dx
(1 – x2) = t ⇒ -2x dx = dt ⇒ -x dx = dt
I = sin-1 x – ∫ \(\frac{d t}{\sqrt{t}}\)
= sin-1 x + ∫ t–\(\frac{1}{2}\)
= sin-1 x + \(\frac{x}{\sqrt{1-x^{2}}}\) + c
விடை :
(2) sin-1 x – \(\sqrt{1-x^{2}}\) + c
Question 17.
∫ \(\frac{d x}{e^{x}-1}\) =
(1) log |ex| – log |ex – 1| + c
(2) log |ex| + log |ex – 1| + c
(3) log |ex – 1| – log|ex| + c
(4) log|ex + 1| – log |ex| + c
குறிப்பு :
ex –ஆல் பெருக்கி வகுக்க
I = ∫ \(\frac{d x}{e^{x}-1}\)
t = ex; dt = ex dx என்க
I = ∫ \(\frac{d t}{t(t-1)}\)
\(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}\) என்க
⇒ 1 = A(t – 1) + Bt
t = 1 எனில், B = 1
t = 0 எனில், A = -1
∴ \(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\)
I = ∫\(\frac{d t}{t(t-1)}\)
= ∫ \(-\frac{1}{t}\) dt + ∫ \(\frac{1}{t-1}\) dt
= – log |t| + log |t – 1| + c
= log |ex – 1| – log |ex| + c
விடை :
(3) log |ex – 1| – log|ex| + c
Question 18.
∫ e-4x cos x dx =
(1) \(\frac{e^{-4 x}}{17}\) [4 cos x – sin x] + c
(2) \(\frac{e^{-4 x}}{17}\) [- 4 cos x + sin x] + c
(3) \(\frac{e^{-4 x}}{17}\) [4 cos x + sin x] + c
(4) \(\frac{e^{-4 x}}{17}\) [- 4 cos x – sin x] + c
குறிப்பு :
I = ∫ e-4x cos x dx
∫ eax cos bx dx = \(\frac{e^{\alpha x}}{a^{2}+b^{2}}\) [a cos bx + b sin bx] + c
இங்கு a = -4, b = 1
∴ I = \(\frac{e^{-4 x}}{(-4)^{2}+1^{2}}\) × [-4 cos x + sin x] + c
= \(\frac{e^{-4 x}}{17}\) [-4 cos x + sin x] + c
விடை :
(2) \(\frac{e^{-4 x}}{17}\) [- 4 cos x + sin x] + c
Question 19.
∫ \(\frac{\sec ^{2} x}{\tan ^{2} x-1}\) dx =
(1) 2 log \(\left|\frac{1-\tan x}{1+\tan x}\right|\) + c
(2) log \(\left|\frac{1+\tan x}{1-\tan x}\right|\) + c
(3) \(\frac{1}{2}\) log \(\left|\frac{\tan x+1}{\tan x-1}\right|\) + c
(4) \(\frac{1}{2}\) log \(\left|\frac{\tan x-1}{\tan x+1}\right|\) + c
குறிப்பு :
I = ∫ \(\frac{\sec ^{2} x}{\tan ^{2} x-1}\) dx
t = tan x ⇒ dt = sec2 x dx என்க
∴ I = ∫ \(\frac{d x}{t^{2}-1^{2}}\)
= \(\frac{1}{2(1)} \log \left|\frac{t-1}{t+1}\right|\) + c
[∵ ∫ \(\frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + c]
= \(\frac{1}{2}\) log \(\left|\frac{\tan x-1}{\tan x+1}\right|\) + c
விடை :
(4) \(\frac{1}{2}\) log \(\left|\frac{\tan x-1}{\tan x+1}\right|\) + c
Question 20.
∫ e-7x sin 5x dx =
(1) \(\frac{e^{-7 x}}{74}\) [-7 sin 5x – 5 cos 5x] + c
(2) \(\frac{e^{-7 x}}{74}\) [7 sin 5x + 5 cos 5x] + c
(3) \(\frac{e^{-7 x}}{74}\) [7 sin 5x – 5 cos 5x] + c
(4) \(\frac{e^{-7 x}}{74}\) 9-7 sin 5x + 5 cos 5x] + c
குறிப்பு :
I = ∫ e-7x sin 5x dx
∫ eax sin bx dx = \(\frac{e^{a x}}{a^{2}+b^{2}}\) [a sin bx – b cos bx] + c
இங்கு a = -7; b = 5
∴ I = \(\frac{e^{-7 x}}{(-7)^{2}+5^{2}}\) [- 7 sin 5x – 5 cos 5x] + c
= \(\frac{e^{-7 x}}{74}\) [- 7 sin 5x – 5 cos 5x] + c
விடை :
(1) \(\frac{e^{-7 x}}{74}\) [-7 sin 5x – 5 cos 5x] + c
Question 21.
∫ x2 e\(\frac{x}{2}\) dx is
(1) x2 e\(\frac{x}{2}\) – 4x e\(\frac{x}{2}\) – 8 e\(\frac{x}{2}\) + c
(2) 2 x2 e\(\frac{x}{2}\) – 8x e\(\frac{x}{2}\) – 16 x\(\frac{x}{2}\) + c
(3) 2 x2 e\(\frac{x}{2}\) – 8x e\(\frac{x}{2}\) + 16 e\(\frac{x}{2}\) + c
(4) \(\frac{e^{\frac{x}{2}}}{2}-\frac{x e^{\frac{x}{2}}}{4}+\frac{e^{\frac{x}{2}}}{8}\) + c
குறிப்பு :
பெர்னோலியின் சூத்திரத்தைப் பயன்படுத்தி
∫ u dv = uv – u’v1 + u”v2
u = x2 u’ = 2x;
dv = e\(\frac{x}{2}\);
v = \(\frac{e^{\frac{x}{2}}}{\frac{1}{2}}\) = 2 e\(\frac{x}{2}\)
v1 = 2 . \(\frac{e^{\frac{x}{2}}}{\frac{1}{2}}\) = 4 e\(\frac{x}{2}\)
u” = 2; v2 = 4 . \(\frac{e^{\frac{x}{2}}}{\frac{1}{2}}\) = 8 e\(\frac{x}{2}\)
∴ I = x2 . (2 e\(\frac{x}{2}\)) – 2x (4 e\(\frac{x}{2}\)) + 2 (8 e\(\frac{x}{2}\)) + c
= 2x2 e\(\frac{x}{2}\) – 8xe\(\frac{x}{2}\) + 16e\(\frac{x}{2}\) + c
விடை :
(3) 2 x2 e\(\frac{x}{2}\) – 8x e\(\frac{x}{2}\) + 16 e\(\frac{x}{2}\) + c
Question 22.
∫ \(\frac{x+2}{\sqrt{x^{2}-1}}\) dx =
(1) \(\sqrt{x^{2}-1}\) – 2 log|x + \(\sqrt{x^{2}-1}\)| + c
(2) sin-1 x – 2 log|x + \(\sqrt{x^{2}-1}\)| + c
(3) 2 log|x + \(\sqrt{x^{2}-1}\)| – sin-1 x + c
(4) \(\sqrt{x^{2}-1}\) + 2 log|x + \(\sqrt{x^{2}-1}\)| + c
குறிப்பு :
I = ∫ \(\frac{x+2}{\sqrt{x^{2}-1}}\) dx
I = ∫ \(\frac{x}{\sqrt{x^{2}-1}}\) dx + ∫ \(\frac{2}{\sqrt{x^{2}-1}}\) dx
t = x2 – 1 ⇒ dt = 2x dx
⇒ \(\frac{dt}{2}\) = x dx
∴ I = \(\frac{1}{2}\) ∫ \(\frac{d t}{\sqrt{t}}\) + 2 ∫ \(\frac{d x}{\sqrt{x^{2}-1}}\)
= \(\frac{1}{2}\) 2√t + 2 log |x + \(\sqrt{x^{2}-1}\)|
= \(\sqrt{x^{2}-1}\) + 2 log |x + \(\sqrt{x^{2}-1}\)| + c
[∵ ∫ \(\frac{d x}{\sqrt{x^{2}-a^{2}}}\) = log |x + \(\sqrt{x^{2}-a^{2}}\)|]
விடை :
(4) \(\sqrt{x^{2}-1}\) + 2 log|x + \(\sqrt{x^{2}-1}\)| + c
Question 23.
∫ \(\frac{1}{x \sqrt{(\log x)^{2}-5}}\) dx =
(1) log |x + \(\sqrt{x^{2}-5}\)| + c
(2) log |log x + \(\sqrt{\log x-5}\)| + c
(3) log |log x + \(\sqrt{(\log x)^{2}-5}\)| + c
(4) log |log x – \(\sqrt{(\log x)^{2}-5}\)| + c
குறிப்பு :
I = ∫ \(\frac{1}{x \sqrt{(\log x)^{2}-5}}\) dx
t = log x
⇒ dt = \(\frac{1}{x}\) dx
I = ∫ \(\frac{d t}{\sqrt{t^{2}-5}}\)
= ∫ \(\frac{d t}{\sqrt{t^{2}-(\sqrt{5})^{2}}}\)
= log|t + \(\sqrt{t^{2}-5}\)| + c = log|log x + \(\sqrt{(\log x)^{2}-5}\)| + c
விடை :
(3) log |log x + \(\sqrt{(\log x)^{2}-5}\)| + c
Question 24.
∫ sin √x dx =
(1) 2(-√x cos √x + sin √x) + c
(2) 2(-√x cos √x – sin √x) + c
(3) 2(-√x sin √x – cos √x) + c
(4) 2(-√x sin √x + cos √x) + c
குறிப்பு :
I = ∫ sin √x dx
t = √x ⇒ dt = \(\frac{1}{2 \sqrt{x}}\) dx என்க
⇒ dt = \(\frac{1}{2t}\) dx
⇒ 2t dt = dx
∴ I = ∫ sin t(2t dt) = 2 ∫ t sin t dt
dv = sin t
u = t; v = – cos t;
u’ = 1; v1 = – sin t
பெர்னோலியின் சூத்திரத்தைப் பயன்படுத்தி
I = 2[uv – u’v1] = 2[-t cos t – 1(- sin t)]
= 2[-t cos t + sin t] + c
= 2[-√x cos √x + sin √x] + c [∵ t = √x]
விடை :
(1) 2(-√x cos √x + sin √x) + c
Question 25.
∫ e√x dx =
(1) 2√x(1 – e√x) + c
(2) 2√x(e√x – 1) + c
(3) 2e√x(1 – √x) + c
(4) 2e√x(√x – 1) + c
குறிப்பு :
I = ∫ e√x dx
√x = t; \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2√x dt = 2t dt
I = ∫ et (2t dt) = 2 ∫ t et dt
u = t; u’ = 1; dv = et; v = et; v1 = et
பெர்னோலியின் சூத்திரத்தைப் பயன்படுத்தி
I = 2[uv – u’v1]
=2[tet – 1 . et + c] = 2et(t – 1) + c
= 2e√x(√x – 1) + c [∵ t = √x]
விடை :
(4) 2e√x(√x – 1) + c