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Let 1 - tan² x = t 

-2 tan x. sec² x dx = dt

-1/2∫dt/t = -1/2 logt + c

\(\int\frac{tanx\,sec^2x}{(1-tan^2x)}dx\) = -1/2 log|1 - tan²x| + c

Put sin x = t 

⇒ cos x dx = dt 

∴ The given equation becomes

\(\int\frac{dt}{\sqrt{4-t^2}}\)

= sin^-1 t/2 + C

But t = sin x

= sin^-1\((\frac{sinx}2)\) + C

Formula to be used -\(\int\cfrac{dx}{\sqrt{a^2-x^2}}=sin^{-1}\cfrac{x}{a}+c\) where c is the integrating constant

\(\therefore\int\cfrac{dx}{\sqrt{16-x^2}}\)

\(=\int\cfrac{dx}{\sqrt{4^2-x^2}}\)

= \(sin^{-1}\cfrac{x}{4}+c\) , c being the integrating constant

Let x² = t 

2xdx = dt

1/2∫ cos t dt = 1/2 sin t + c

∫ x cosx² dx = 1/2 sinx² + c

Assume,

1 + cotx = t 

d(1 + cotx) = dt 

⇒cosec²x = dt 

Put t and dt in given equation we get,

⇒ ∫\(\frac{dt}{t}\)

= ln|t| + c

But,

t = 1 + cotx 

= ln|1 + cotx| + c.

(i) ∫sec x(sec x + tan x) dx

It can be written as

= ∫sec² x + sec x tan x dx

By integrating w.r.t x

= tan x + sec x + c

(ii) ∫cosec x(cosec x – cot x) dx

It can be written as

= ∫cosec² x – cosec x cot x dx

By integrating w.r.t x

= – cot x + cosec x + c

Let x² + x + 1 = t 

(2x +1)dx = dt

∫ √tdt = 2/3t^3/2 + c = 2/3(x² + x + 1)^3/2 + c

Let log x = 1

1/x dx = dt

∫sec²t dt = tan t + c

\(\int\frac{sec^2(logx)}{x}dx\) = tan(logx) + c

\(\frac{sec^2x}{{cosec^2 x}}\) = tan²x

∫ tan²x dx = ∫ (sec²x - 1)dx

\(\int\frac{sec^2x}{{cosec^2 x}}dx\) = tan x - x + c

Let log x = t

1/xdx = dt

∫sec²tdt = tan t + c

\(\int\frac{sec^2(log\,x)}{x}dx\) = tan (logx) + c

\(\int\frac{5sin 2x-cos x}{9sin^2x+4}dx\) 

\(=\int\frac{5sin2x}{9sin^2x+4}dx-\int\frac{cos x}{9sin^2x+4}dx\)

= \(5\int\frac{dt}{9t+4}-\frac19\int\cfrac{du}{u^2+\frac49}\) (By taking sin²x = t and sin x = u)

= \(\frac59\)log|9t + 4| - \(\frac16\)tan^-1(3u/2) + c

= \(\frac59\) log|9 sin²x + u| - \(\frac16\) tan^-1(\(\frac{3sinx}2\)) + c

(By putting t = sin²x and u = sin x)

We know that,

∫a^xdx = a^x/In a + c

∫2^xdx = 2^x/In 2 + c

We can write ∫ cos³3x dx as:

∫cos3x(cos3x)²dx ∫cos³x(cos²3x)dx and

further as :

= cos3x(1-sin²3x)dx 

= ∫cos3xdx-∫cos3x(sin²3x)dx 

Taking A = ∫cos3xdx 

Solving for A,

A = \(\frac{sin3x}{3}\)

Taking B = ∫cos3x(sin²3x)dx 

In this taking sin3x = t 

Differentiating on both sides we get,

3cos³xdx = dt 

Solving by putting these values we get,

B = \(\int\frac{t^2}{3}\)dt

= \(\frac{t^3}{9}\) + c

Substituting values we get,

B = \(\frac{sin^33x}{9}\) + c

Our final answer is A+B i.e,

\(\frac{sin3x}{3}\) + \(\frac{sin3x}{3}\) + c

We know that,

\(\int{\sqrt{x^2-a^2}}dx\) = \(\frac{x}2\)\({\sqrt{x^2-a^2}}\) - \(\frac{a^2}2\) log | x + \({\sqrt{x^2-a^2}}\) + c

\(\int{\sqrt{x^2-4^2}}dx\) = \(\frac{x}2\)\({\sqrt{x^2-16}}\) - 8 log | x + \({\sqrt{x^2-16}}\) + c

Let \(\sqrt x\) = z

⇒ \(\frac{1}{2\sqrt x}\,dx=dz\) 

⇒ \(\frac{dx}{\sqrt x} =2dz\) 

∴ \(\int \frac{sec^2\sqrt x}{\sqrt{x}}\,dx\) = \(2\int sec^2z\,dz\)

= tan z + C

= 2 tan \(\sqrt{x}+C\)

Let I = ∫cos √x dx 

Put √x = t 

∴ x = t² 

∴ dx = 2t dt

∴ I = ∫(cos t) 2t dt 

= ∫ 2t cos t dt

= 2t ∫cos t dt – ∫[d/dt(2t) ∫cos t dt]dt

= 2t sin t – ∫2 sin t dt 

= 2t sin t + 2 cos t + c 

= 2[√x sin√x + cos√x] + c.

Assume ,

√x = t 

d(√x) = dt

⇒ \(\frac{1}{2\sqrt x}\) dx = dt

⇒ \(\frac{1}{\sqrt x}\) dx = dt

Substituting t and dt

⇒ 2∫ sint dt 

= - 2cost + c 

But,

√x = t 

⇒2 cos(√x) + c.

Assume,

√x = t 

d(√x) = dt 

⇒ \(\frac{1}{2\sqrt x}\)dx = dt

⇒ \(\frac{1}{\sqrt x}\) = dt

Substituting t and dt 

⇒2 ∫ sec²t dt 

= 2 tan t + c 

But,

√x = t 

⇒2 tan(√x) + c.

Assume,

√x = t 

d(√x) = dt

⇒ \(\frac{1}{2\sqrt x}\) dx = dt

⇒ \(\frac{1}{\sqrt x}\) dx = 2dt

Substituting t and dt

⇒2 \(\int\) cost dt

= 2sint + c 

But,

√x = t 

⇒2 sin(√x) + c.

Given :

∫(sec²x + cosec²x) dx

By Splitting, we get,

∫(sec²x dx + ∫cosec²x dx

By applying the formula,

∫(sec²x dx = tanx

∫cosec²x dx = - cotx

⇒tan x – cot x + c

Given,

∫secx (secx + tanx) dx

=∫ (sec²x + secx . tanx) dx 

= tanx + secx + c

Given,

∫ tan³xsec²xdx

let tan x = t 

Differentiating on both sides we get, 

sec²x dx = dt 

Substituting above equation in ∫tan³x sec²x dx we get, 

= ∫t³ dt

= \(\frac{t^4}{4}\)+ c

= \(\frac{tan^4x}{4}\) + c

Given, 

∫ e^x sec x( 1 + tan x) dx 

= ∫ e^x (sec x + sec x tan x) dx 

= e^x sec x + c 

∵∫e^x (f(x) +f'(x)) dx = e^x f(x)+c

Assume,

sin2x + tanx - 5 = t

d(tanx + sin2x - 5) = dt 

(2cos2x + sec²x) dx = dt 

Put t and dt in given equation we get

⇒ \(\int\frac{dt}{t}\)

= In|t| + c

But,

t = sin2x + tanx - 5 

= ln|sin2x + tanx - 5| + c.

Given,

∫ tan⁶x sec²x dx

let tan x = t 

Differentiating on both sides we get, 

sec²x dx = dt 

Substituting above equation in ∫ tan³x sec²x dx we get, 

=∫t⁶ dt

= \(\frac{t^7}{7}\) + c

= \(\frac{tan^7x}{7}\) + c

Given,

∫ tanx sec³x dx 

= ∫ (tanx secx) sec²x dx 

Let secx = t 

Differentiating on both sides we get, 

tanx secx dx = dt 

Substituting above equation in ∫ tanx sec³x dx we get, 

= ∫t² dt

= \(\frac{t^3}{3}+c\)

= \(\frac{sec^3x}{3}\) + c

Let cosx = t 

Differentiating on both sides we get, 

- sinx dx = dt 

Substituting above equation in ∫cos⁴x sinx dx we get, 

= ∫- t⁴ dt

= \(-\frac{t^5}{5}\) + c

= \(-\frac{cos^5x}{5}\) + c

Let sin x = t 

Differentiating on both sides we get, 

cosx dx = dt 

Substituting above equation in ∫sin³x cosx dx we get, 

= ∫t³ dt

= \(\frac{t^4}{4}\) + c

= \(\frac{sin^4x}{4}\) + c

Assume,

x + log(sinx) = t 

d(x + log(sinx)) = dt 

1 + \(\frac{cos\,x}{sin\,x}\)dx = dt

(1 + cot)dx = dt 

Put t and dt in given equation we get,

⇒∫\(\frac{dt}{t}\)

= In|t| + c

But,

t = x + log(sinx) 

= ln| x + log(sinx) | + c.

Given,

 ∫ e^{log sinx} cosx dx

= ∫sin x cos x dx 

(∵e^logx = x)

Let sin x = t

Differentiating on both sides we get, 

Cos x dx = dt 

Substituting above equations in given equation we get, 

= ∫t dt

= \(\frac{t^2}{2}\) + c

= \(\frac{sin^2x}{2}\) + c

Consider ∫ e^{3 logx} x⁴

e^{3 logx} = \(e^{log\,x^3}\)

= x³ 

∫ e^{3 logx} x⁴ 

= ∫ x³x⁴dx 

= ∫ x⁷ dx

= \(\frac{x^8}{8}\) + c

Assume,

log(sinx) = t 

d(log(sinx)) = dt 

⇒ \(\frac{cosx}{sinx}\)dx = dt

⇒ cot x dx = dt 

Substituting the values oft and dt we get

⇒ \(\int\)t dt

⇒ \(\frac{t^2}{2}\) + c

But,

t = log(sinx)

⇒ \(\frac{log(sin\,x)^2x}{2}\) + c.

Let x + log sin x = t 

Differentiating it on both sides we get, 

(1+cot x) dx = dt - i 

Given that,

 ∫(1 + cotx)/(x + log sinx)dx

Substituting i in above equation we get, 

= \(\int\frac{dt}{t}\)

= log t + c 

= log(x + log sin x ) + c

Here the denominator is already factored. 

So let,

∫ (2x+1)/(x+1)(x-2)dx

= \(\frac{A}{x+1}+\frac{B}{x-2}\) ...(i)

⇒ \(\int\frac{2x+1}{(x+1)(x-2)}\) = \(\frac{A(x-2)+B(x+1)}{(x+1)(x-2)}\)

⇒ 2x + 1 = A(x – 2) + B(x + 1)……(ii)

We need to solve for A and B. 

One way to do this is to pick values for x which will cancel each variable. 

Put x = 2 in the above equation, we get

⇒ 2(2) + 1 = A(2 – 2) + B(2 + 1)

⇒ 3B = 5

⇒ B = \(\frac{5}{3}\)

Now 

Put x = – 1 in equation (ii), we get 

⇒ 2( – 1) + 1 = A(( – 1) – 2) + B(( – 1) + 1) 

⇒ – 3A = – 1

⇒ A = \(\frac{1}{3}\)

We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. 

We get,

\(\int[\frac{A}{x+1}+\frac{B}{x-2}]\) dx

⇒ \(\int[\frac{\frac{1}{3}}{x+1}+\frac{\frac{5}{3}}{x-2}]\) dx

Split up the integral,

⇒ \(\frac{1}{3}\)\(\int[\frac{1}{x+1}]\)dx + \(\frac{5}{3}\)\(\int[\frac{1}{x-2}]\)dx

Let substitute u = x + 1 

⇒ du = dx and z = x – 2 

⇒ dz = dx, 

So the above equation becomes,

⇒ \(\frac{1}{3}\)\(\int[\frac{1}{u}]\)du + \(\frac{5}{3}\)\(\int[\frac{1}{z}]\)dz

On integrating we get,

⇒ \(\frac{1}{3}\)log|u| + \(\frac{5}{3}\)log|z| + c

Substituting back, we get

⇒ \(\frac{1}{3}\)log|x + 1| + \(\frac{5}{3}\)log|x - 2| + c

Note : the absolute value signs account for the domain of the natural log function (x>0). 

Hence,

\(\int\frac{2x+1}{(x+1)(x-2)}\) dx = \(\frac{1}{3}\)log|x + 1| + \(\frac{5}{3}\)log|x - 2| + c

Given,

 \(\int\frac{log\,x^x}{x}\) dx

= \(\int\frac{x\,log\,x}{x}\)dx

= ∫log x 

= x log x-x + c

Given,

∫ sec²(7 - 4x)dx

Let 7 – 4x = t 

Differentiating on both sides we get, 

-4 dx = dt

dx = \(-\frac{1}{4}\) dt

Substituting it in ∫ sec²(7 - 4x)dx we get,

= \(\int-\frac{1}{4}\) sec²t dt

= tan t + c 

= tan (7 - 4x) + c

Given,

∫ 1/(x(1+logx))dx

Let 1+log x = t

⇒ \(\frac{d}{dx}\) (1 + logx) = dt

⇒ \(\frac{1}{x}\) dx = dt

= \(\int\frac{1}{t}\) dt

= logt +c 

= log (1+logx)+c

Given, 

f’(x) = 8x³ – 2x and f(2) = 8 

On integrating the given equation, we have

\(\int\)f'(x)dx =  \(\int\)(8x³ -2x) dx

We know,

 \(\int\)f'(x)dx = f(x)

⇒  f(x) =  \(\int\)(8x³ -2x) dx

⇒  f(x) =  \(\int\)(8x³ -2x) dx - \(\int\)2xdx

⇒  f(x) =  8\(\int\)(x³ dx - 2\(\int\)xdx

Recall \(\int\)xⁿdx = \(\frac{x^{n+1}}{n+1}\) + c

⇒  f(x) =  8(\(\frac{x^{3+1}}{3+1}\)) - 2(\(\frac{x^{1+1}}{1+1}\)) + c

⇒  f(x) =  8(\(\frac{x^{4}}{4}\)) - 2(\(\frac{x^{2}}{2}\)) + c

⇒ f(x) = 2x⁴ – x² + c 

On substituting x = 2 in f(x), we get 

f(2) = 2(2⁴) – 2² + c 

⇒ 8 = 32 – 4 + c 

⇒ 8 = 28 + c 

∴ c = –20 

On substituting the value of c in f(x), we get 

f(x) = 2x⁴ – x² + (–20) 

∴ f(x) = 2x⁴ – x² – 20 

Thus, 

f(x) = 2x⁴ – x² – 20

Assume,

2sinx + cosx  = t 

d(2sinx + cosx) = dt 

(2cosx - sinx)dx = dt 

Put t and dt in given equation we get

⇒ ∫\(\frac{dt}{t}\)

= In|t| + c

But,

t = 2sinx + cosx 

= ln|2sinx + cosx| + c.

Assume,

x + tan^-1x = t 

d(x + tan^-1x) = dt

⇒ 1 + \(\frac{1}{x^2+1}\) = dt

⇒ \(\frac{2+x^2}{x^2+1}\) = dt

Substituting t and dt

⇒ \(\int\)5^tdt

⇒ \(\frac{5^t}{log\,5}\) + c  

But,

t = x + tan^-1x

⇒ \(\frac{5^{x+tan^{-1}x}}{log\,5}\) + c.

Given,

∫√((1-x)/x) dx

Let,

\(\sqrt x\) = t

\(\frac{d}{dx}(\sqrt x)\) = dt

\(\frac{1}{2\sqrt x}\) dx = dt

dx = 2t dt

Now,

\(\int\frac{\sqrt {1-t^2}}{t}\) 2t dt

= 2\(\int\sqrt{1-t^2}\) dt

Consider, 

t = sin k 

dt = cos k dk

= 2\(\int\sqrt{1-sin^2k}\) .cosk dk

= 2\(\int\sqrt{cos^2k}\) .cosk dk

= 2 ∫cos²k dk

= ∫2 cos²k dk 

=∫cos 2k-1 dk 

[since, cos 2x = 2cos²x-1]

= \(\frac{sin\,2k}{2}\) - k + c

= \(\frac{2sink\,cosk}{2}\) - k + c

= t cos(sin^-1t) - 2sin^-1t + 2c 

=√x cos(sin^-1√x )- 2sin^-1√x + 2c

Given,

∫ (x+cos6x)/(3x² + sin6x)dx

Let 3x² + sin6x = t

⇒ \(\frac{d}{dx}\) (3x² + sin6x) = dt

⇒ 6x + cos 6x. 6 = dt

⇒ x + cos 6x = \(\frac{dt}{6}\)

Substituting the values,

= \(\int\frac{1}{6t}\) dt

= \(\frac{1}{6}\)log t +c 

= \(\frac{1}{6}\)log(3x² + sin6x) +c

Correct answer is (a) \(\frac{tan^{m + 1}X}{m + 1} + c\)

Correct answer is (a) \(\frac{e^x}{x} + c\)

Given:

\(\int3^xdx\)

\(\int a^xdx=\frac{a^x}{In\,a}+c\)

\(\int 3^xdx=\frac{3^x}{In\,3}+c\)

To find: \(\int\frac{dx}{(x^2+16)}\)

Formula Used: \(\int\cfrac{dx}{a^2+x^2}=\frac{1}{a}tan^{-1}(\frac{x}{a})+c\)

Rewriting the given equation,

\(\Rightarrow\)\(\int\frac{dx}{4^2+x^2}\)

Here a = 4

\(\Rightarrow\)\(\cfrac{1}{4}\times tan^{-1}(\frac{x}{4})+c\)

Therefore,

\(\int\cfrac{dx}{(x^2+16)}=\frac{1}{4}\times tan^{-1}(\frac{x}{4})+c\)\(\)

To find: \(\int\cfrac{cosx}{(1+sin^2x)}dx\)

Formula Used: \(\int\cfrac{dx}{a^2+x^2}=\cfrac{1}{a}tan^{-1}(\cfrac{x}{a})+C\)

Let y = sin x … (1)

Differentiating both sides, we get

dy = cos x dx

Substituting in given equation,

\(\Rightarrow\)\(\int\cfrac{dy}{1+y^2}\)

⇒ tan^-1 y

From (1),

⇒ tan^-1 (sin x)

Therefore,

\(\int\cfrac{cosx}{(1+sin^2x)}dx=tan^{-1}(sinx)+C\)

Correct answer is (b) \(\frac{2}{3} (x + 1) ^{\frac{3}{2}}+ c\)

\(\int\frac{dx}{\sqrt{x^2-a^2}}\) = log | x + \(\sqrt{x^2-a^2}\)|

\(\int\frac{dx}{\sqrt{x^2-4^2}}\) = log | x + \(\sqrt{x^2-16}\)|

\(\int\frac{3e^x}{e^{2x} + 5e^x + 6}dx\)

= 3\(\int\frac{t}{t^2 + 5t + 6}dt\)

put e^x = t

= 1\(\int\frac{3tdt}{(t + 2)(t + 3)} = \int\frac{A}{t + 2} + \frac{B}{t + 3}dt\)

Let 36 = A(t + 3) + B(t + 2) 

put t = -2 - 6 = A(1) ⇒ A = -6 

put t = -3, - 9 = A(0) + B(-1) ⇒ B = 9

∴ \(\int\frac{3e^x}{(e^x)^2 + 5e^x + 6} = \int\frac{-6}{t + 2} + \frac{9}{t + 3}dt\)

= -6log(t + 2) + 9log(t + 3) + C = -6log(e^x + 3) + 9 log(e^x + 3) + c