Samacheer Class 11 Maths - Integral Calculus

(i) \(\frac{1}{\sin ^{2} x}\)
\(\frac{1}{\sin ^{2} x}\) = ∫cosec 2 x dx
= – cot x + c
(ii) \(\frac{\tan x}{\cos x}\)
\(\frac{\tan x}{\cos x}\) = ∫sec x tan x dx
= sec x + c
(iii) \(\frac{\cos x}{\sin ^{2} x}\)
(iv) \(\frac{1}{\cos ^{2} x}\)
\(\frac{1}{\cos ^{2} x}\) = ∫sec 2 x dx
= tan x + c

(i) 12 3
∫12 3 dx = 12 3 ∫dx = 12 3 x + c
(ii) \(\frac{x^{24}}{x^{25}}\)
(iii) e x
∫e x dx = e x + c

(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) cosec 2 (5x – 7)
[∫cosec 2 (ax + b) dx = – \(\frac{1}{a}\) cot (ax + b) + c]
∫cosec 2 (5x – 7) dx = – \(\frac{1}{5}\) cot (5x – 7) + c

(i) e 3x – 6
[∫e ax+b dx = \(\frac{1}{a}\) e ax + b + c]
∫e 3x – 6 dx = \(\frac{1}{3}\) e 3x – 6 + c
(ii) e 8 – 7x
[∫e ax+b dx = \(\frac{1}{a}\) e ax + b + c]
∫e 8 – 7x = \(\frac{1}{-7}\) e 8 – 7x + c
∫e 8 – 7x = \(-\frac{1}{7}\) e 8 – 7x + c
(iii) \(\frac{1}{6-4 x}\)

(i) sec 2 \(\frac{x}{5}\)
[∫sec 2 (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

\(\int f^{\prime}(x) d x=\int(4 x-5) d x\)
f(x) = \(\frac{4 x^{2}}{2}\) – 5x + c
f(x) = 2x 2 – 5x + c
But f(2) = 1
2(2) 2 – 5(2) + c = 1
8 – 10 + c = 1
c = 3
Thus, f(x) = 2x 2 – 5x + 3

Given f’ (x) = 9x 2 – 6x and f(0) = – 3
\(\frac{\mathrm{d} f(x)}{\mathrm{d} x}\) = 9x 2 – 6x
df(x) = (9x 2 – 6x) dx
∫ df(x) = ∫ (9x 2 – 6x)
∫ df(x) = ∫ 9x 2 dx – ∫ 6x dx
f(x) = 9 ∫x 2 dx – 6 ∫ x dx
f(x) = 9 × \(\frac{x^{3}}{3}\) – 6 \(\frac{x^{2}}{2}\) + C
f(x) = 3x 3 – 3x 2 + c —— (1)
f(o) = – 3
(1) ⇒ f(0) = 3 × – 3. 0 2 + c
– 3 = 0 – 0 ⇒c = – 3
Substituting in equation (1) we get
f(x) = 3x 3 – 3x 2 – 3
f(x) = 3(x 3 – x 2 – 1)

\(\)
f'(x) = \(\) – 6x + c
f'(x) = 6x 2 – 6x + c
But f'(1) = 5
6(1) 2 – 6(1) + c = 5
c = 5
f” (x) = 6x 2 – 6x + 5
\(\int f^{\prime \prime}(x) d x=\int\left(6 x^{2}-6 x+5\right) d x\)
f(x) = \(\frac{6 x^{3}}{3}-\frac{6 x^{2}}{2}\) + 5x + c
f(x) = 2x 3 – 3x 2 + 5x + c
But f(1) = 30
2(1) 3 – 3(1) 2 + 5(1) + c = 30
2 – 3 + 5 + c = 30
c = 26
f(x) = 2x 3 – 3x 2 + 5x + 26

Initial velocity of the ball = 39.2 in / s
Let s be the distance of the ball from the ground at time t.
u = 39.2m/s.
s = ut – \(\frac{1}{2}\). gt 2 where g = 9.8 m / sec.
s = 39.2 t – \(\frac{1}{2}\) × (9.8) t 2
s = 39.2 t – 49 t 2
(i) how long will it take for the ball to strike the ground?
For the ball to strike the ground, s = 0
∴ 39.2 t – 4.9 t 2 = 0
t(39.2 – 4.9 t) = 0
39.2 – 4.9 t = 0 since t ≠ 0
4.9 t = 39.2
t = \(\frac{39.2}{4.9}\) = 8
t = 8 sec.
(ii) The speed with which will it strike the ground?
At t = 8; (1) ⇒ v = -9.8(8) + 39.2
= -78.4 + 39.2
= -39.2
∴ The speed with which the ball will strike the ground is = 39.2 m/s
(iii) At maximum height v = 0
⇒ (2) ⇒ -9.8 t + 39.2 = 0
t = 4 sec
⇒ (3) ⇒ x = \(-\frac{9.8 \times 16}{2}\) + 39.2 × 4
= -78.4 + 156.8 = 78.4 m/s

(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate?
(ii) From Sunday to Thursday there are 4 days. Substituting t = 4 in equation (2) we get
∴ The anticipated area of the wound on Thursday = 1.5 sq. cm.

∫(2x – 5) (3x + 4x) dx
= ∫(72 x + 8x 2 – 180 – 20x) dx
= ∫ 72 x dx + ∫82x 2 dx – ∫180 dx – ∫2o x dx
= 72∫x dx + 8∫x 2 dx – 180 ∫dx – 20 ∫x dx

∫cot 2 x + tan 2 x
= ∫(cosec 2 x – 1 + sec 2 x – 1) dx
= ∫(cosec 2 x + sec 2 x – 2) dx
= ∫cosec 2 x dx + ∫sec 2 x dx – ∫2 dx
= – cot x + tan x – 2x + c
= tan x – cot x – 2x + c

= 3 ∫ cosec 2 x dx + 4 ∫ cot x cosec x. dx
= 3 ∫ cosec 2 x dx + 4 ∫ cosec x cot x dx
= 3 × – cot x + 4 × – cosec x + c
= – 3 cot x – 4 cosec x + c

[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

∫e x log a e x = ∫e log a x. 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

∫ sin 5 x cos 3 x dx = ∫ sin 5 x cos 2 x. cos x dx
= ∫ sin 5 x (1 – sin 2 x). cos x dx
= ∫ (sin 5 x – sin 7 x). cos x dx
= ∫ sin 5 x cos x dx – ∫ sin 7 x cos x dx
put u = sin x
du = cos x dx

(ii) x sin 3x
(iii) 25 x e -5x
(iv) x sec x tan x
∫ x sec x tan x dx
u = x, u’ = 1, u” = 0
dv = sec x tan x dx, v = ∫ sec x tan x dx = sec x
v 1 = ∫ v dx = ∫ sec x dx = log |sec x + tan x|
v 2 = ∫ v 1 dx = ∫ log |sec x + tan x| dx
∫ u dv = uv – u’ v 1 + u” v 2 – u”’ v 3 + ………….
∫x sec x tan x dx = x sec x – 1 × log |sec x + tan x| + 0 × ∫ log |x sec x tan x| + c
∫x sec x tan x dx = x sec x – log |sec x + tan x| + c

(ii) 27 x 2 e 3x
(iii) x 2 cos x
(iv) x 3 sin x
∫ x 3 sin x
u = x 3, u’ = 3x 2, u” = 6x, u”’ = 6, u’ v = 0
dv = sin x dx ⇒ v = ∫ sin x dx = – cosx
v 1 = ∫v dx = ∫- cos x dx = – ∫ cos x dx = – sin x
v 2 = ∫ v 1 dx = ∫ – sin x dx = – ∫ sin x dx = – (- cos x) = cos x
v 3 = ∫ v 2 dx = ∫ cos x dx = sin x
v 4 = ∫ v 3 dx = ∫ sin x dx = – cos x
∫ u dv = uv – u’ v 1 + u” v 2 – u”’ v 3 + u’ v v 4 – ………..
∫x 3 sin x dx = x 3 (- cos x) – 3x 2 (- sin x) + 6x (cos x) – 6 sin x + 0 (- cos x) + c
= -x 3 cos x + 3x 2 sin x + 6x cos x – 6 sin x + c

Let I = ∫e x (tan x + log sec x) dx
Take f(x) = log sec x
f'(x) = \(\frac{1}{\sec x}\) sec x tan x
f'(x) = tan x
[∫e x [f (x) + f (x)] dx = e x f(x) + c]
∴ I = e x log |sec x| + c

Let I = \(\mathrm{I}=\int e^{x}(\sec x+\sec x \tan x) d x\)
Take f(x) = sec x
f ‘ (x) = sec x tan x
This is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\) = e x f(x) + c
= e x sec x + c

Let 2x – 3 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x 2 + 4x – 12) + B
2x – 3 = A (2x + 4) + B
2x – 3 = 2Ax + 4A + B
2A = 2 ⇒ A = 1
4A + B = – 3
4 × 1 + B – 3 ⇒ B = – 3 – 4 = – 7
2x – 3 = 1 (2x + 4) – 7
(ii) \(\frac{5 x-2}{2+2 x+x^{2}}\)
Let 5x – 2 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x 2 + 2x + 2) + B
5x – 2 = A (2x + 2) + B
5x – 2 = 2Ax + 2A + B
2A = 5 ⇒ A = \(\frac{5}{2}\)
2A + B = -2
2 × \(\frac{5}{2}\) + B = -2 ⇒ B = -2 – 5 = -7
5x – 2 = \(\frac{5}{2}\) (2x + 2) – 7
Put x 2 + 2x + 12 = t
(2x + 2) dx = dt
Put x + 1 = u
dx = du
(iii) \(\frac{3 x+1}{2 x^{2}-2 x+3}\)
Let 3x + 1 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (2x 2 – 2x + 3) + B
3x + 1 = A (4x – 2) + B
3x + 1 = 4Ax – 2A + B

Let 2x + 1 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (9 + 4x – x 2 ) + B
2x + 1 = A(4 – 2x) + B
2x + 1 = 4A – 2Ax + B
-2A = 2 ⇒ A = -1
4A + B = 1 ⇒ 4 (-1) + B = 1
B = 1 + 4 = 5
B = 5
2x + 1 = – 1 (4 – 2x) + 5
Put 9 + 4x – x 2 = t
(4x – 2) dx = dt
Put x – 2 = u
dx = du
(ii) \(\frac{x+2}{\sqrt{x^{2}-1}}\)
Let x + 2 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x 2 – 1) + B
x + 2 = A (2x) + B
2A = 1 ⇒ A = \(\frac{1}{2}\)
B = 2
(iii) \(\frac{2 x+3}{\sqrt{x^{2}+4 x+1}}\)
Let 2x + 3 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x 2 + 4x + 1) + B
2x + 3 = A (2x + 3) + B
2A = 2 ⇒ A = 1
4A + B = 3
4 × 1 + B = 3 ⇒ B = 3 – 4
B = -1
(2x + 3) = (2x + 4) – 1
Put x 2 + 4x + 1 = t
(2x + 4) dx = dt
Put x + 2 = u
dx = dt

(1) ∫ (f(x) 2 dx
Explaination:
Given ∫ f (x) dx = g (x) + c
\(\frac{\mathrm{d}}{\mathrm{d} x}\) ∫ f(x)dx = \(\frac{\mathrm{d}}{\mathrm{d} x}\) (g(x) + c)
∫ \(\frac{\mathrm{d}}{\mathrm{d} x}\) (f(x)) dx = g'(x)
∫ d(f(x)) = g'(x)
f(x) = g(x)
∴ ∫ f(x) g'(x) dx = ∫ f(x) f(x) dx
= ∫ [f(x)] 2 dx

(4) \(\frac{2 x^{3}}{3}\) – x 2 + x + c
Explaination:
Given ∫ f'(x) e x 2 dx = (x – 1)e x 2 + c
Differentiating both sides with respect to x we have

(1) y = x + \(\frac{4}{x}\) + 3
Explaination:
Given, this curve passes through the point (2, 7)
∴ 7 = 2 + \(\frac{4}{2}\) + c
7 = 2 + 2 + c
c = 7 – 4 = 3
∴ The required equation is
y = x + \(\frac{4}{x}\) + 3

(3)
Explaination:
∫sin 3 dx
sin 3x = 3 sin x – 4 sin 3 x
4 sin 3 x = 3 sin x – sin 3x
sin 3 x = \(\frac{1}{4}\)(3 sin x – sin 3x)

(1) x 2 sin x + 2x cos x – 2 sin x + c
Explaination:
\(\int x^{2} \cos x d x\)
By Bernoullis formula dv = cosxdx
u = x 2 v = sinx
u’ = 2x v 1 = -cos x
u” = 2 v 2 = -sinx
= uv – u’v 1 + u”v 2
= x 2 sin x + 2x cos x – 2 sin x + c