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From question, the integration given is:

\(I = \smallint \left( {\frac{{{{\left( {{\rm{si}}{{\rm{n}}^n}\;\theta - {\rm{sin\;}}\theta } \right)}^{\frac{1}{n}}}{\rm{cos\;}}\theta }}{{{\rm{si}}{{\rm{n}}^{n + 1}}\theta }}} \right)d\theta\)

On putting, sin θ = t

⇒ cos θ dθ = dt

Now,

\(\Rightarrow I = \smallint \frac{{{{\left( {{t^n} - t} \right)}^{1/n}}}}{{{t^{n + 1}}}}dt\)

\(\Rightarrow I = \smallint \frac{{{{\left[ {{t^n}\left( {1 - \frac{t}{{{t^n}}}} \right)} \right]}^{1/n}}}}{{{t^{n + 1}}}}dt\)

\(\Rightarrow I = \smallint \frac{{t{{\left( {1 - \frac{1}{{{t^{n - 1}}}}} \right)}^{1/n}}}}{{{t^{n + 1}}}}dt\)

\(\Rightarrow I = \smallint \frac{{{{\left( {1 - 1/{t^{n - 1}}} \right)}^{1/n}}}}{{{t^n}}}dt\)

On putting,

\(\Rightarrow 1 - \frac{1}{{{t^{n - 1}}}} = u\)

∴ 1 – t-(n-1) = u

\(\Rightarrow \frac{{\left( {n - 1} \right)}}{{{t^n}}}dt = du\)

\(\therefore \frac{{dt}}{{{t^n}}} = \frac{{du}}{{n - 1}}\)

On substituting in integral,

\(\Rightarrow I = \smallint \frac{{{u^{1/n}}du}}{{n - 1}} = \frac{{{u^{\frac{1}{n} + 1}}}}{{\left( {n - 1} \right)\left( {\frac{1}{n} + 1} \right)}} + C\)

\(\Rightarrow I = \frac{{n{{\left( {1 - \frac{1}{{{t^{n - 1}}}}} \right)}^{n + 1}}}}{{\left( {n - 1} \right)\left( {n + 1} \right)}} + C\)

\(\left[ {u = 1 - \frac{1}{{{t^{n - 1}}}}{\rm{\;and\;}}t = {\rm{sin}}\theta } \right]\)

\(\Rightarrow I=\frac{n{{\left( 1-\frac{1}{\text{si}{{\text{n}}^{n-1}}\theta } \right)}^{\frac{n+1}{n}}}}{{{n}^{2}}-1}+C\)

\(I = \smallint \frac{{{\rm{sin}}\left( {\frac{{5{\rm{x}}}}{2}} \right)}}{{{\rm{sin}}\left( {\frac{{\rm{x}}}{2}} \right)}}{\rm{dx}}\)

\(= \smallint \frac{{{\rm{sin}}\left( {2x + \frac{x}{2}} \right)}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)

\(= \smallint \frac{{\sin 2x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)

\(= \smallint \frac{{2\sin x\cos x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)

[∵ sin 2x = 2 sin x cos x]

\(= \smallint \frac{{2\sin x\cos x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)

\(= \smallint \frac{{2 \times 2\sin \frac{x}{2}\cos \frac{x}{2}\cos x\cos \frac{x}{2} + \cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)

\(\left[ \because{\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}} \right]\)

\(= \smallint \frac{{4\sin \frac{x}{2}\cos \frac{x}{2}\cos x\cos \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}} + \frac{{\cos 2x\sin \frac{x}{2}}}{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{\rm{dx}}\)

\(= \smallint \left[ {4\cos x{{\cos }^2}\frac{x}{2} + \cos 2x} \right]{\rm{dx}}\)

\(= \smallint \left[ {4\cos x\left( {\frac{{1 + \cos x}}{2}} \right) + \cos 2x} \right]{\rm{dx}}\)

\(\left[ \because{\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} } \right]\)

= ∫[2 cos x (1 + cos x) + cos 2x] dx

= ∫[2 cos x + 2 cos2 x + cos 2x] dx

\(= \smallint \left[ {2\cos x + 2\left( {\frac{{1 + \cos 2x}}{2}} \right) + \cos 2x} \right]{\rm{dx}}\)

= ∫[2 cos x + 1 + cos 2x + cos 2x] dx

= ∫[2 cos x + 1 + 2 cos 2x] dx

\(= 2{\rm{sinx}} + {\rm{x}} + \frac{{2{\rm{sin\;}}2{\rm{x}}}}{2} + {\rm{c}}\)

Concept:

Integral properties: Consider a function f(x) defined on x.

• \(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},\;\;\;f\left( {\rm{x}} \right) = f\left( { - x} \right)}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f\left( {\rm{x}} \right) = - f\left( { - x} \right)} \end{array}} \right.\)

Calculation:

Given:

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^3}{\rm{dx}} = 0{\rm{\;}}\)

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^2}{\rm{dx}} = \frac{2}{3}\)

\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^3}{\rm{dx}} = 0{\rm{\;}}\)

Let f(x) = x3

F(-x) = -x3

⇒ f(-x) = - f(x)

∴ f(x) is an odd function.

We know that, if f(x) is odd function then,\(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 0\)

So, b = -a ----(1)

Now,\(\mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} {{\rm{x}}^2}{\rm{dx}} = \frac{2}{3}\)

\(\Rightarrow \left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]_{\rm{a}}^{\rm{b}} = \frac{2}{3}\)

⇒ b3 – a3 = 2

⇒ -a3 – a3 = 2 [Using (1)]

⇒ a3 = -1

⇒ a = -1

So, b = 1

Let, \(I = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\rm{se}}{{\rm{c}}^{\frac{2}{3}}}x\cdot{\rm{cose}}{{\rm{c}}^{\frac{4}{3}}}\;xdx\)

\( = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{{1.dx}}{{{\rm{co}}{{\rm{s}}^{\frac{2}{3}}}x\cdot{\rm{si}}{{\rm{n}}^{\frac{4}{3}}}x}}\)

\(\left[\because {\frac{1}{{cos\theta }} = sec\;\theta \;and\;\frac{1}{{sin\theta }} = cosec\theta } \right]\)

\( = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{{1dx}}{{{\rm{co}}{{\rm{s}}^2}x\cdot{\rm{ta}}{{\rm{n}}^{\frac{4}{3}}}x}} = \mathop \smallint \nolimits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \frac{{{\rm{se}}{{\rm{c}}^2}xdx}}{{{\rm{ta}}{{\rm{n}}^{\frac{4}{3}}}x}}\)

Let tanx = u

\(I = \mathop \smallint \nolimits_{\frac{1}{{\sqrt 3 }}}^{\sqrt 3 } {u^{ - \frac{4}{3}}}\;du\)

\( = \frac{{\left[ {{u^{ - \frac{1}{3}}}} \right]_{\frac{1}{{\sqrt 3 }}}^{\sqrt 3 }}}{{ - \frac{1}{3}}} = \frac{{3\left[ {{u^{ - \frac{1}{3}}}} \right]_{\frac{1}{{\sqrt 3 }}}^{\sqrt 3 }}}{{ - 1}}\)

\( = - 3\left[ {{3^{ - \;\frac{1}{6}}} - \frac{1}{{{3^{\frac{{ - 1}}{6}}}}}} \right] = - 3\left( {{3^{\frac{{ - 1}}{6}}} - {3^{\frac{1}{6}}}} \right) = 3\left( {{3^{\frac{1}{6}}} - {3^{\frac{{ - 1}}{6}}}} \right) = \left( {{3^{\frac{7}{6}}} - {3^{\frac{5}{6}}}} \right)\)

CONCEPT:

• \(\smallint \sin x\;dx = \; - \cos x + C\)
• \(\smallint {e^x}\;dx = {e^x} + C\)
• \(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

CALCULATION:

Given:\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = - \frac{1}{a}\cos 2x - \frac{b}{3}\;{e^{3x}} + C\)

As we know that,\(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

⇒\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = \smallint \sin 2x\;dx - \;\smallint 4{e^{3x}}\;dx\)

As we know that,\(\smallint \sin x\;dx = \; - \cos x + C\)and\(\smallint {e^x}\;dx = {e^x} + C\)

⇒\(\smallint \sin 2x\;dx + \;\smallint 4{e^{3x}}\;dx = - \frac{1}{2} cos \ 2x - \frac{4}{3} e^{3x} + C\)

⇒\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = - \frac{1}{2} cos \ 2x - \frac{4}{3} e^{3x} + C\)

Now, by comparing the above equation with\(\smallint \left( {\sin 2x - 4{e^{3x}}} \right)\;dx = - \frac{1}{a}\cos 2x - \frac{b}{3}\;{e^{3x}} + C\)we get,

⇒ a = 2 and b = 4

⇒ a + b = 6

Hence, correct option is 1.

CONCEPT:

• \(\smallint {x^n}\;dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\)
• \(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

CALCULATION:

Given:\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{a} + bx + C\)

As we know that,\(\smallint \left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \;\smallint f\left( x \right)\;dx \pm \;\smallint g\left( x \right)\;dx\)

⇒\(\smallint \left( {{x^2} - 1} \right)\;dx = \smallint {x^2}\;dx - \smallint dx \)

As we know that,\(\smallint {x^n}\;dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\)

⇒\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{3} - x + C\)

Now, by comparing the above equation with\(\smallint \left( {{x^2} - 1} \right)\;dx = \frac{{{x^3}}}{a} + bx + C\)we get

⇒ a = 3 and b = - 1

⇒ a + 3b = 0

Hence, correct option is 2.

From question, the equation given is:

\(\mathop \smallint \nolimits_\alpha ^{\alpha + 1} \frac{{{\rm{d}}x}}{{\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}} = {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left( {\frac{9}{8}} \right)\)

Let’s assume the integral as ‘I’,

\(\Rightarrow I = \mathop \smallint \nolimits_\alpha ^{\alpha + 1} \frac{{{\rm{d}}x}}{{\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}\)

On putting, t = (x + α),

\(\Rightarrow \frac{{dt}}{{dx}} = 1 \Rightarrow dt = dx\)

For limits:

⇒ x = α + 1 ⇒ t = 2α + 1

⇒ x = α ⇒ t = 2α

On substituting the integral becomes,

\(\Rightarrow I = \mathop \smallint \nolimits_{2\alpha }^{2\alpha + 1} \frac{{{\rm{d}}t}}{{\left( t \right)\left( {t + 1} \right)}}\)

On breaking the above integral,

\(\Rightarrow I = \mathop \smallint \nolimits_{2\alpha }^{2\alpha + 1} \left( {\frac{1}{t} - \frac{1}{{t + 1}}} \right)dt\)

\(\left[ {\smallint \frac{1}{x}dx = {\rm{ln}}\left| x \right| + C} \right]\)

\(\Rightarrow I=\left[ \ln \left| t \right|-\ln \left| t+1 \right| \right]_{2\alpha }^{2\alpha +1}\)

\(\because \left[ \text{ln}\left( \frac{x}{y} \right)=\text{ln}\left( x \right)-\text{ln}\left( y \right) \right]\)

\(\Rightarrow I = \left[ {\ln \left( {\frac{t}{{t + 1}}} \right)} \right]_{2\alpha }^{2\alpha + 1}\)

On direct substitution,

\(\Rightarrow I = \left[ {\ln \left( {\frac{{2\alpha + 1}}{{2\alpha + 1 + 1}}} \right)} \right] - \left[ {\ln \left( {\frac{{2\alpha }}{{2\alpha + 1}}} \right)} \right]\)

From question,

\(\Rightarrow \left[ {\ln \left( {\frac{{2\alpha + 1}}{{2\alpha + 1 + 1}}} \right)} \right] - \left[ {\ln \left( {\frac{{2\alpha }}{{2\alpha + 1}}} \right)} \right] = {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left( {\frac{9}{8}} \right)\)

\(\Rightarrow \left[ {\ln \left( {\frac{{2\alpha + 1}}{{2\alpha + 2}}} \right)} \right] - \left[ {\ln \left( {\frac{{2\alpha }}{{2\alpha + 1}}} \right)} \right] = {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left( {\frac{9}{8}} \right)\)

\(\because \left[ \text{ln}\left( \frac{x}{y} \right)=\text{ln}\left( x \right)-\text{ln}\left( y \right) \right]\)

\(\Rightarrow \ln \left( \frac{\left( \frac{2\alpha +1}{2\alpha +2} \right)}{\left( \frac{2\alpha }{2\alpha +1} \right)} \right)=\text{lo}{{\text{g}}_{\text{e}}}\left( \frac{9}{8} \right)\)

\(\Rightarrow \ln \left( \frac{2\alpha +1}{2\alpha +2}\times \frac{2\alpha +1}{2\alpha } \right)=\text{lo}{{\text{g}}_{\text{e}}}\left( \frac{9}{8} \right)\)

\(\Rightarrow \ln \left( \frac{{{\left( 2\alpha +1 \right)}^{2}}}{2\alpha \left( 2\alpha +2 \right)} \right)=\text{lo}{{\text{g}}_{\text{e}}}\left( \frac{9}{8} \right)\)

On cancelling ln and loge,

\(\Rightarrow \frac{{{\left( 2\alpha +1 \right)}^{2}}}{2\alpha \left( 2\alpha +2 \right)}=\frac{9}{8}\)

∵ [(a + b)2 = a2 + b2 + 2ab]

\(\Rightarrow \frac{4{{\alpha }^{2}}+1+4\alpha }{4{{\alpha }^{2}}+4\alpha }=\frac{9}{8}\)

⇒ 8(4α2 + 1 + 4α) = 9(4α2 + 4α)

⇒ 32α2 + 8 + 32α = 36α2 + 36α

⇒ 36α2 + 36α – 32α2 – 8 – 32α = 0

⇒ 4α2 + 4α – 8 = 0

⇒ 4 (α2 + α – 2) = 0

⇒ α2 + α – 2 = 0

⇒ α2 – α + 2α – 2 = 0

⇒ α(α – 1) + 2(α – 1) = 0

⇒ (α + 2)(α – 1) = 0

∴ α = -2, 1

The given equation is:

\(\text{I}=\int \left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{4}}+\text{x}} \right)\text{dx}\)

Now, dividing x2 in numerator and denominator,

\(\Rightarrow \text{I}=\int \left( \frac{\left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{2}}} \right)}{\left( \frac{{{\text{x}}^{4}}+\text{x}}{{{\text{x}}^{2}}} \right)} \right)\text{dx}\)

\(\Rightarrow \text{I}=\int \left( \frac{2\text{x}-{{\text{x}}^{-2}}}{{{\text{x}}^{2}}+{{\text{x}}^{-1}}} \right)\text{dx}\)

\(\left[ \because \int \left( \frac{\text{{f}'}\left( \text{x} \right)}{\text{f}\left( \text{x} \right)} \right)\text{dx}={{\log }_{\text{e}}}\left( \text{f}\left( \text{x} \right) \right)+\text{C} \right]\)

⇒ I = loge(x2 + x-1) + C

\(\Rightarrow \text{I}={{\log }_{\text{e}}}\left( {{\text{x}}^{2}}+\frac{1}{\text{x}} \right)+\text{C}\)

Calculation:

Given:

\(\displaystyle\int\dfrac{\sin x}{\cos (x-a)}dx=Ax+B\log |sec(x-a)|+c\)

\(Let = \int\frac{sin\ x}{sin (x - a)}dx\)

Put t = x - a

Differentiating w. r. t. x

⇒\(\frac{dt}{dx}=\frac{d(x-a)}{dx}\)

⇒\(\frac{dt}{dx}=1\)

⇒\(dx = dt\)

⇒\(\int\frac{sin(t + a)}{sin\ t}dt\)

⇒\(\int\frac{sin\ t\ cos\ a \ +\ cos\ t \ sin\ a}{sin\ t}dt \) \([sin (A+ B)= sin A cosB + cosA sinB]\)

⇒\(\int\frac{sint \ cosa }{sint}+\frac{cost\ sina}{sint}dt\)

⇒\(\int{cos\ a \ dt }+\int{cot \ t\ sina\ dt}\)

⇒\(cos a \int{dt + sin a}+ \int {cot \ t\ dt}\) (since\(sin a, cosa\)are constant)

⇒\(cos a ×t + sin a\ log |sint |+c\)

⇒ putting back t = x - a

⇒\({(x - a)}cos a+sina[sin(x-a)]+ c\)

⇒\(sin a\ log |sin (x-a)+ xcos a- a cosa + c\)

⇒\(sin a\ log|sin(x-a)|+xcosa + c_1\) \((c_1 = - acosa + c)\)

Concept:

\(\displaystyle\int \dfrac{dt}{t}\)= log t + c

\(\displaystyle∫{\dfrac{1-\tan x}{1+\tan x}}\ dx\)

Let's convert tan x into sin x and cos x.

\(\Rightarrow \displaystyle\int \dfrac{1-\dfrac{\sin x}{\cos x}}{1+\dfrac{\sin x}{\cos x}}dx\) (∵\(\tan x = \dfrac{\sin x}{\cos x}\) )

\(\Rightarrow \displaystyle\int \dfrac{\dfrac{\cos x-\sin x}{\cos x}}{\dfrac{\cos x + \sin x}{\cos x}}dx\)

\(\Rightarrow \displaystyle\int \dfrac{\cos x-\sin x}{{\cos x + \sin x}}\)dx ----(1)

Let cos x + sin x = t

Differentiating thisequation on both sides,

∴\(\dfrac{\mathrm{d}(\cos x + \sin x)}{\mathrm{d} x}=\dfrac{\mathrm{d} t}{\mathrm{d} x}\)

⇒(cos x - sin x)dx = dt

Subsitutingthis value in equation (1),

⇒ log t + c

Subsituting the value of t in this equation,

⇒ log (cosx + sin x) + c

Hence,\(\displaystyle∫{\dfrac{1-\tan x}{1+\tan x}}\ dx\)=log (cos x + sin x) + c

Concept:

\(\int sec^2x\ dx = tanx\)

\(\int sec\ x \ tan\ x \ dx= sec\ x\)

Calculation:

\(\displaystyle\int \dfrac{1}{1+ sinx } dx= \displaystyle\int \dfrac{1}{1+ sinx } dx \times \dfrac{1-sinx}{1- sinx}\)

\(=\displaystyle\int \dfrac{1- sinx }{1- sin^2x } dx= \displaystyle\int \dfrac{1- sinx}{cos^2 } dx\)

\(= \displaystyle\int\left[ \dfrac{1}{cos^2x }- \dfrac{sinx}{cos^2 x} dx\right]\)

\(= \displaystyle\int(sec^2 x - sec x. tan x ) dx\)

= tan x - sec x + c