Let I = \(\int { { e }^{ ax } } \) cos bx
Take u = cos bx ; dv =e^ax dx
du = -b sin bx dx ; V = \(\frac { { e }^{ ax } }{ a } \)
Applying intergration by parts we get,
I = uv - \(\int { vdu } \)
= \(\frac { { e }^{ ax } }{ a } \) cos bx + \(\int { \frac { { e }^{ ax } }{ a } } \).b sin bc dx
I = \(\frac { { e }^{ ax } }{ a } \) cos bx + \(\frac { b }{ a } \) I₁
Consider I₁ = \(\int { { e }^{ ax } } \) sin bx dx
Take u = sin bx ; dv = e^ax dx
du = b cps bx dx; V = \(\frac { { e }^{ ax } }{ a } \) v =  \(\frac { { e }^{ ax } }{ a } \)
Applying intergration by parts we get,
I₁ = uv - \(\int { vdu } \) = \(\frac { { e }^{ ax }sinbx }{ a } -\int { \frac { { e }^{ ax } }{ a } } \) b.cos bx dx
= \(\frac { { e }^{ ax }sinbx }{ a } -\frac { b }{ a } \) \(\int { { e }^{ ax } } \) cos bx dx
I₁ = \(\frac { { e }^{ ax }sinbx }{ a } -\frac { b }{ a } \)I
Substituting I₁ in (1) we get,
I = \(\frac { { e }^{ ax } }{ a } \) cos bx + \(\frac { b }{ a } \) \(\left( \frac { { e }^{ ax } }{ a } sinbx-\frac { b }{ a } I \right) \)
= \(\frac { { e }^{ ax } }{ a } cosbx+\frac { b }{ { a }^{ 2 } } sinbx-\frac { { b }^{ 2 } }{ { a }^{ 2 } } I\)
\(\Rightarrow\)  \(\left( 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) I=\frac { ae^{ ex }cosbx+be^{ ax }sinbx }{ { a }^{ 2 } } \)
\(\Rightarrow\) \(\left( \frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 } } \right) I=\frac { ae^{ ex }cosbx+be^{ ax }sinbx }{ { a }^{ 2 } } \)
\(\therefore \int e^{a x} \cos b x d x\) = \({e^{ax}\over a^2+b^2}[a \ cos \ bx + b \ sin \ bx]+c\)

\(\int { \left( 1+{ x }^{ 2 } \right) ^{ -1 } } dx=\int { \frac { 1 }{ 1+{ x }^{ 2 } } } dx\)
= tan^-1 x + c

\(\int 12^3 d x =\int k d x \text { where } k=12^3 \)
\(=k x+c=12^3 x+c . \)

\(\int {1\over 1+x^2}dx=tan^{-1}x+c\)

\(\int { {1\over cos ^2 \ x}dx }=\int {sec^2 x dx =tan \ x+c} \)

We know that \(\int { x^n \ dx }={x^{n+1}\over n+1} +c,n\neq -1.\)
Putting n = 10, we get
\(\int { x^{10} \ dx }={x^{10+1}\over 10+1} +c={x^{11}\over 11}+c\)

Let I =\(\int{1\over (x+2)^2+1}dx=\int {1\over (x-2)^2+1^2}dx\)
Putting x - 2 = t \(\Rightarrow\)dx = dt
Thus, I =\(\int {1\over t^2+1^2}dt=tan^{-1}(t)+c=tan^{-1}(x-2)+c\)

\( \int \frac{1+\cos 4 x}{\cot x-\tan x} d x =\int \frac{2 \cos ^2 2 x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x} d x} \)
\(=\int \frac{2 \sin x \cos x \cos ^2 2 x}{\cos ^2 x-\sin ^2 x} d x \)
\(=\int \frac{\sin 2 x \times \cos ^2 2 x}{\cos 2 x} d x \)
\(=\int \sin 2 x \cos 2 x d x\)
\(=\int \frac{1}{2} \sin 4 x d x \)
\( =\frac{1}{2}\left(\frac{-\cos 4 x}{4}\right)+c \)
\(=\frac{-1}{8} \cos 4 x+c \)

\(\int { \left( 2x-5 \right) } \left( 36+4x \right) dx=\int { \left( 72x+{ 8x }^{ 2 }-180-20x \right) dx } =\int { \left( { 8x }^{ 2 }+52x-180 \right) } dx\)
\(=8\left(\frac{x^3}{3}\right)+52\left(\frac{x^2}{2}\right)-180 x+c\)
= \({8x^3\over 3}+26x^2-180x+c\)

\(\int e^{x \log 2} e^x d x=\int e^{\log 2^x} e^x d x=\int 2^x e^x d x\)
\(=\int(2 e)^x d x=\frac{(2 e)^x}{\log (2 e)}+c\)

\(\int(tan \ x+cot \ x)^2dx=\int [tan^2x+2tan \ x \ cot \ x+cot^2\ x]dx\)
\(=\int [(sec^2x-1)+2+(cosec^2 x-1)]dx\)
\(=\int (sec^2x+cosec^2 x)dx\)
= tan x + (-cot x) + c
= tan x - cot x + c.

Given: \(f^{\prime}(x)=4 x-5\)
\( \therefore \int f^{\prime}(x) d x =\int(4 x-5) d x\)
\( \therefore f(x) =4\left(\frac{x^2}{2}\right)-5(x)+c\)
\(=2 x^2-5 x+c \)

\( \int e^x d x =e^x+c\)
\(\therefore \int e^{3 x-6} d x =\frac{e^{3 x-6}}{3}+c\)

\(\int{1\over (3x-2)}dx={1\over3}log|(3x-2)|+c\)

Let \(\frac{x+1}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}\)
Multiplying both sides by (x + 2)(x + 3)
x + 1 =  A(x + 3) + B(x + 2) ............(1)
Putting x = -2 in (1)
-2 + 1 = A(-2 + 3) + 0
-1 = A(1) ⇒ A = -1
Putting x = -3 in (1)
-3 +1 = 0 + B(-3 + 2)
-2 = B(-1) ⇒ B = 2
\(\therefore \frac{x+1}{(x+2)(x+3)}=\frac{-1}{x+2}+\frac{2}{x+3}\)
\(\therefore \int \frac{x+1}{(x+2)(x+3)} d x=\int\left[\frac{-1}{x+2}+\frac{2}{x+3}\right] d x\)
\( =-\int \frac{1}{x+2} d x+2 \int \frac{1}{x+3} d x\)
\(=-\log |x+2|+2 \log |x+3|+c \)
\(=2 \log |x+3|-\log |x+2|+c \)

The rate of change of height h with respect to time t is the derivative of h with respect to t.
Therefore,\({dh\over dt}={18\over \sqrt{t}}=18t^{-{1\over2}}\)
So, to get a general expression for the height, integrating the above equation with respect to t.
\(h=\int {18t^{-{1\over 2}}}dt=18(2t^{1\over2})+c=36\sqrt{t}+c\)
Given that when t = 0, the height h = 5 cm.
\(5=0+c \Rightarrow c=5\)
\(h=36\sqrt{t}+5\).
(i) To find the height of the tree after 4 years.
When t = 4 years,
\(h=36\sqrt{t}+5\Rightarrow h=36\sqrt{4}+5=77\)
The height of the tree after 4 years is 77 cm .
(ii) When h = 149 cm
\(h=36 \sqrt{t}+5 \Rightarrow 149=36 \sqrt{t}+5\)
\(\sqrt{t}=\frac{149-5}{36}=4 \Rightarrow t=16\)
Thus after 16 years the height of the tree will be 149 cm.

\(=\int ({15\over \sqrt{5x-4}}-8cot (4x+2)cosec(4x+2))dx\)
\(=15\int {1\over \sqrt{5x-4}}dx-8\int cot (4x+2)cosec(4x+2)dx\)
\(=15({1\over 5})(2\sqrt{5x-4})-8({1\over4})(-cosec (4x+2)+c\)
\(=6\sqrt{5x-4}+2cosec(4x+2)+c\)

\(\text { Let } \sqrt{x}=t\)
\(\text { Then } x=t^{2}\)
\(\therefore d x =2 t d t \)
\(\therefore \int e^{\sqrt{x}} d x =\int e^{t} \times 2 t d t \)
\(=2 \int t e^{t} d t \)
Applying Bernoulli's formula
\(=2\left[t\left(e^{t}\right)-1\left(e^{t}\right)\right]+c \)
\(=2(t-1) e^{t}+c \)
\(\text { Putting } t=\sqrt{x}\)
\(=2(\sqrt{x}-1) e^{\sqrt{x}}+c\)
 

\(\int \frac{x+2}{\sqrt{x^{2}-1}} d x =\int\left(\frac{x}{\sqrt{x^{2}-1}} d x+\frac{2}{\sqrt{x^{2}-1}}\right) d x \)
\(=\sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+c \)

Applying Bernoulli's formula
\(\int x^{2} e^{x / 2} d x =x^{2}\left(\frac{e^{x / 2}}{1 / 2}\right)-2 x\left(\frac{e^{x / 2}}{1 / 2 \times 1 / 2}\right)+2\left(\frac{e^{x / 2}}{1 / 2 \times 1 / 2 \times 1 / 2}\right)+c \)
\(=2 x^{2} e^{x / 2}-8 x e^{x / 2}+16 e^{x / 2}+c \)
 

\(\int \frac{d x}{e^{x}-1} =\int \frac{1}{e^{x}\left(1-\frac{1}{e^{x}}\right)} d x \)
\(=\int \frac{e^{-x}}{1-e^{-x}} d x \)
\(=\log \left|1-e^{-x}\right|+c \)
\(=\log \left|1-\frac{1}{e^{x}}\right|+c \)
\(=\log \left|\frac{e^{x}-1}{e^{x}}\right|+c \)
\(=\log \left|e^{x}-1\right|-\log \left|e^{x}\right|+c \)

\(\int \sqrt{\frac{1-x}{1+x}} d x =\int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \)
\(=\int \sqrt{\frac{(1-x)^{2}}{1^{2}-x^{2}}} d x \)
\(=\int \frac{1-x}{\sqrt{1-x^{2}}} d x \)
\(=\int\left(\frac{1}{\sqrt{1-x^{2}}}-\frac{x}{\sqrt{1-x^{2}}}\right) d x \)
\(=\sin ^{-1} x-\left(-\sqrt{1-x^{2}}\right)+c \)
\(=\sin ^{-1} x+\sqrt{1-x^{2}}+c \)

\(\int \frac{x^{2}+\cos ^{2} x}{x^{2}+1} \operatorname{cosec}^{2} x d x \)
\(=\int \frac{x^{2}+\left(1-\sin ^{2} x\right)}{x^{2}+1} \times \frac{1}{\sin ^{2} x} d x \)
\(=\int \frac{\left(x^{2}+1\right)-\sin ^{2} x}{\left(x^{2}+1\right) \sin ^{2} x} d x \)
\(=\int\left(\frac{1}{\sin ^{2} x}-\frac{1}{x^{2}+1}\right) d x \)
\(=\int\left(\operatorname{cosec}^{2} x-\frac{1}{x^{2}+1}\right) d x \)
\(=-\cot x-\tan ^{-1} x+c \)

\(\sin ^{8} x-\cos ^{8} x =\left(\sin ^{4} x-\cos ^{4} x\right)\left(\sin ^{4} x+\cos ^{4} x\right) \)
\(=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right) \)
\(\times \left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right] \)
\(=(1)(-\cos 2 x)\left(1-2 \sin ^{2} x \cos ^{2} x\right) \)
\( \therefore \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}=-\cos 2 x \)
\( \therefore \int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x=\frac{-\sin 2 x}{2}+c \)

\(\int \tan ^{-1} \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x =\int \tan ^{-1} \sqrt{\tan ^{2} x} d x \\ \)
\(=\int \tan ^{-1}(\tan x) d x \)
\(=\int x d x=\frac{x^{2}}{2}+c \)

\(\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x =\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \)
\(=\int \frac{x^{2}\left(x^{4}-x^{3}\right)}{x^{4}-x^{3}} d x \)
\(=\int x^{2} d x=\frac{x^{3}}{3}+c \)

\(\text { Given: } \int f(x) d x=g(x)+c\)
\(\therefore f(x) =g^{\prime}(x) \)
\(\therefore \int f(x) g^{\prime}(x) d x =\int[f(x)]^{2} d x \)