Mathematics Test - 6

Concept:

\(\rm \int \frac{dx}{\sqrt {a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}+C\)

Calculation:

Given: \(\rm \int \frac{dx}{\sqrt {9-x^{2}}}=sin^{-1}\frac{x}{b}+C\)

Using the formula, 

\(\rm \int \frac{dx}{\sqrt {a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}+C\)

\(\rm \Rightarrow \int \frac{dx}{\sqrt {9-x^{2}}}=\int \frac{dx}{\sqrt {3^{2}-x^{2}}}=sin^{-1}\frac{x}{3}+C\)     -----(1)

∵ It is given that, \(\rm \int \frac{dx}{\sqrt {9-x^{2}}}=sin^{-1}\frac{x}{b}+C\)-     ----(2)

On comparing (1) and (2) we get b = 3.

Hence, the correct answer is option 2.

Given, f(x) = 12x ³ – 15x² + 64

Antiderivative of f(x) is F(x) + C, where C is a constant.

\(F\left( x \right) = \smallint 12{{\rm{x}}^3}{\rm{}}-{\rm{}}15{{\rm{x}}^2}{\rm{}} + {\rm{}}64dx = \frac{{12{x^4}}}{4} - \frac{{15{x^3}}}{3}+ 64x + C = 3{x^4} - 5{x^3} + 64x + C\)

Concept:

\(\rm \int \sqrt {x^2+a^2}\ dx=\frac{x}{2}\sqrt {x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt {x^2+a^2}\right|+C\).

Calculation:

Let I = \(\rm \int \sqrt {x^2+2x+5}\ dx\)

= \(\rm \int \sqrt {x^2+2x+1+4}\ dx\)

= \(\rm \int \sqrt {(x+1)^2+2^2}\ dx\)

Using \(\rm \int \sqrt {x^2+a^2}\ dx=\frac{x}{2}\sqrt {x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt {x^2+a^2}\right|+C\), we get:

= \(\rm \frac{x+1}{2}\sqrt {x^2+2x+5}+\frac{2^2}{2}\log\left|x+1+\sqrt {x^2+2x+5}\right|+C\)

= \(\rm \frac{1}{2}\left( {x + 1} \right)\sqrt {{x^2} + 2x + 5} + 2\log \left| {x + 1 + \sqrt {{x^2} + 2x + 5} } \right| + C\)

Concept:

Partial Fraction:

Calculation:

Here we have to find the value of \(\smallint \frac{{dx}}{{{e^x} - 1}}\)

Let e^x = t and by differentiating ex = t with respect to x we get

⇒ e^x dx = dt or dx = dt/e^x = dt/t

\(⇒ \smallint \frac{{dx}}{{{e^x} - 1}} = \;\smallint \frac{{dt}}{{t\left( {t - 1} \right)}}\)

Let \(\frac{1}{{t\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{t - 1}}\)

⇒ 1 = A (t - 1) + B t ---------(1)

By putting t = 0 on both the sides of (1) we get A = - 1

By putting t = 1 on both the sides of (1) we get B = 1

\(\Rightarrow \frac{1}{{t\left( {t - 1} \right)}} = \frac{{ - 1}}{t} + \frac{1}{{t - 1}}\)

\(\Rightarrow \smallint \frac{{dt}}{{t\left( {t - 1} \right)}} = \; - \;\smallint \frac{{dt}}{t} + \;\smallint \frac{{dt}}{{t - 1}}\)

As we know that \(\smallint \frac{{dx}}{x} = \log \left| x \right|\; + C\)  where C is a constant

 \(\Rightarrow \smallint \frac{{dt}}{{t\left( {t - 1} \right)}} = \; - \log \left| t \right| + \log \left| {t - 1} \right| + C\)

\(= \log \left| {\frac{{t - 1}}{t}} \right| + C\)

By substituting  ex = t in the above equation we get

\(\Rightarrow \smallint \frac{{dx}}{{{e^x} - 1}} = \log \left| {\frac{{{e^x} - 1}}{{{e^x}}}} \right| + C\)

Concept:

sin 2x = 2sin x cos x

∫(1/x)dx = log x + c

∫tanx dx = sec²x + c  

Calculation:

Let I = \(\smallint \frac{2}{{\sin 2{\rm{x}}.\log \left( {{\rm{tanx}}} \right)}}\)         ....(1)

Take log (tan x) = t

\(\rm \frac1 {\tan x}(se{c^2}x)dx = dt\)

⇒ \( \frac{{{\rm{cosx}}}}{{{\rm{sinx}}.{\rm{\;co}}{{\rm{s}}^2}{\rm{x}}}}{\rm{\;dx}} = {\rm{dt}}\)

⇒ \( \frac{1}{{{\rm{sinx}}.{\rm{cosx}}}}{\rm{dx}} = {\rm{dt}}\)

⇒ dx = sin x.cos x dt

Putting the value of log (tan x) and dx in equation (i)

Now, I = \(\rm \smallint \frac{2}{{2sinx.cosx.t}}\;sinx.cosx\;dt\)

= ∫ \(\rm \frac 1 t\)dt

= log t + c

= log [log(tan x) ]+ c

\(I = \smallint {e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx\)

\(Let\;I' = \smallint {e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx = \smallint {e^x}f\left( x \right)dx + \smallint {e^x}f'\left( x \right)dx\)

Integration by Parts of first term:

\(\smallint f\left( x \right)g\left( x \right)dx\; = \;f\left( x \right)\smallint g\left( x \right)dx - \smallint \left[ {f'\left( x \right)\smallint g\left( x \right)dx} \right]dx\)

Where f is first function and g is second function.

Preference order generally adopted for the selection of the first function: ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

f = f(x); g = e^x

\(I' = \;\left\{ {f\left( x \right)\smallint {e^x}dx - \smallint f'\left( x \right)\left[ {\smallint {e^x}dx} \right]dx} \right\} + \smallint {e^x}f'\left( x \right)dx\)

\(I' = \;\left\{ {{e^x}f\left( x \right) - \smallint {e^x}f'\left( x \right)dx} \right\} + \smallint {e^x}f'\left( x \right)dx = {e^x}f\left( x \right)\)

\(\therefore \smallint {e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx = {e^x}f\left( x \right) + c\)

Here \(f\left( x \right) = \;\frac{1}{x};f'\left( x \right) = - \frac{1}{{{x^2}}}\)

\(I = \smallint {e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx = \frac{{{e^x}}}{x} + c\)

Concept:

sin²x + cos²x = 1

\(\rm \int sin x dx = -cosx \)

\(\rm \int cos x dx = sin x \)

Calculation:

We have \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{sin^{2}x + cos^{2}x - 2sinx cosx} )dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{(sinx - cosx)^{2}})dx\) = A sinx + B cosx + C

⇒ \(\rm \int (|(sinx - cosx)|)dx\) = A sinx + B cosx + C     ----(i)

If 0 < x < \(\frac{\pi}{4}\), then sinx < cosx

⇒ |sinx - cosx| = -sinx + cosx    -----(ii)

Now from (i) and (ii), we get 

⇒ \(\rm \int (-\space sinx + cosx) \space dx\) = A sinx + B cosx + C

⇒ cosx + sinx + C = A sinx + B cosx + C

On comparing A = 1, B = 1 and C = 0

Hence, A + B - 2 = 0 is correct.

Calculation:

Given:\(I_1 = \displaystyle\int_e^{e^2} \dfrac{dx}{\log x}\) and \(I_2 = \displaystyle\int_1^2 \dfrac{e^x}{x} dx\)

⇒ \(I_1 = \displaystyle\int_e^{e^2} \dfrac{dx}{\log x}\) put log x = z

Such that x = e^z

Such that dx = e^z dz 

when x = e, z = loge

x = e², z = log e² = 2 log e = z 

Such that I₁ = \(\displaystyle\int_{1}^{2}\)(e^z dz) / z =\(\displaystyle\int_{1}^{2}\)(e^x/z) dx = I₂

Such that I₁ = I₂

⇒ I1 - I2 = 0 

Concept:

\(\rm \displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}dx= \sin^{-1}(\frac{x}{a})+c\)

Calculation:

Let I = \(\rm \displaystyle\int \dfrac{1}{\sqrt{9-16x^2}}dx\)

\(= \rm \displaystyle\int \dfrac{1}{4\sqrt{\frac{9}{16}-x^2}}dx\\=\rm \frac{1}{4}\displaystyle\int \dfrac{1}{\sqrt{(\frac{3}{4})^2-x^2}}dx\\=\frac{1}{4} \sin^{-1}[\frac{x}{(\frac{3}{4})}] + c\\=\frac{1}{4} \sin^{-1}(\frac{4x}{3})+c\)

Now, I = \(\frac{1}{4} \sin^{-1}(\frac{4x}{3})+c=α \sin^{-1} (\beta x)+c\)

Therefore, α = \(\dfrac{1}{4} \text{and} \; \beta = \dfrac{4}{3}\)

To Find:

\(\rm \alpha + \dfrac{1}{\beta}= \dfrac{1}{4} + \dfrac{3}{4} \\ =\dfrac{4}{4}\\=1\)

Concept:

\( \rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

To Find: Integral of sec² x with respect to sec x

\(\rm \int \sec^2 x\; d(\sec x)\)

Replace sec x = t, we get

\(= \rm \int t^2 dt\\= \frac{t^3}{3}+c\\=\frac{\sec^3 x}{3}+c\)