Indefinite Integrals MCQ Quiz in मल्याळम - Objective Question with Answer for Indefinite Integrals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• Integration by Parts:

  ∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

• ∫ sin x dx = - cos x + C

Calculation:

Let I = ​​∫ ex(sin x - cos x) dx.

⇒ I = ∫ ex sin x dx - ∫ ex cos x dx

⇒ I = ex ∫ sin x dx - ∫ [e^x ∫ sin x dx] dx - ∫ ex cos x dx

⇒ I = - ex cos x dx + ∫ ex cos x dx - ∫ ex cos x dx + C

⇒ I = - ex cos x + C

As we know, ∫ ex [f(x) + f'(x)]dx = ex f(x) + c

Let f(x) = -cos x

So, f'(x) = sin x

Now, I =  ∫ ex(sin x - cos x) dx

=  ​​∫ ex(- cos x + sin x) dx

= ∫ ex [f(x) + f'(x)]dx

=  ex f(x) + c

= - ex cos x + C

Concept:

Using substitution method in indefinite integral and using formula $\int sinxdx=-cosx+C$

Solution:

$\int e^x.sin{e}^{x}dx$

Substitute u = e^x → $\frac{\mathrm{d}u}{\mathrm{d}x}=e^x\to du=e^{x}dx$

$=\int \sin udu$

$=-\cos u+C$

Undo substitution u = e^x

$=-\cos e^x+C$

∴ The correct option is (2)

Concept:

• sin2x + cos2x = 1  ...(i)
• sin 2x  = 2 sin x cos x   ...(ii)
• sin (x + y) = sin x cos y + sin y cos x
• cos 2x = $2{\cos }^{2}x-1=1-2{\sin }^{2}x$

Calculation:

$\int \frac{1}{\sin x+\cos x}dx$

Multiplying numerator and denominator by $\frac{1}{\sqrt{2}}$,

⇒ $\frac{1}{\sqrt{2}}\int \frac{1}{(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)}dx$ 

⇒ $\frac{1}{\sqrt{2}}\int \frac{1}{(\sin x\cos \frac{\pi }{4}+\sin \frac{\pi }{4}\cos x)}dx$ 

⇒ $\frac{1}{\sqrt{2}}\int \frac{1}{\sin (x+\frac{\pi }{4})}dx$ 

⇒ $\frac{1}{\sqrt{2}}\int \frac{1}{2\sin (\frac{x}{2}+\frac{\pi }{8})\cos (\frac{x}{2}+\frac{\pi }{8})}dx$ 

Multiplying numerator and denominator by $\cos (\frac{x}{2}+\frac{\pi }{8})$,

⇒ $\frac{1}{2\sqrt{2}}\int \frac{{\sec }^2(\frac{x}{2}+\frac{\pi }{8})}{\tan (\frac{x}{2}+\frac{\pi }{8})}dx$ 

Consider $\tan (\frac{x}{2}+\frac{\pi }{8})$ = t ⇒ $\frac{1}{2}{\sec }^2(\frac{x}{2}+\frac{\pi }{8})dx=dt$

⇒ $\frac{1}{\sqrt{2}}\int \frac{dt}{t}$ 

⇒ $\frac{1}{\sqrt{2}}\log (t)$ + c

⇒ $\frac{1}{\sqrt{2}}\log \tan (\frac{x}{2}+\frac{\pi }{8})+c$

∴ The correct answer is option (3).

Concept:

1 + cos 2x = 2cos2 x

1 - cos 2x = 2sin2 x

$\int \cos \mathrm{x}\mathrm{d}\mathrm{x}=\sin \mathrm{x}+\mathrm{c}$

Calculation:

I = $\int \mathrm{s}\mathrm{i}{\mathrm{n}}^2\mathrm{x}\mathrm{d}\mathrm{x}$

= $\int \frac{1-\cos 2\mathrm{x}}{2}\mathrm{d}\mathrm{x}$

= $\frac{1}{2}\int (1-\cos 2\mathrm{x})\mathrm{d}\mathrm{x}$

= $\frac{1}{2}[\mathrm{x}-\frac{\sin 2\mathrm{x}}{2}]+\mathrm{c}$

= $\frac{\mathrm{x}}{2}-\frac{\sin 2\mathrm{x}}{4}+\mathrm{c}$

Concept:

1. Integration by Substitution:

• If the given integration is of the form $\int \mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right){\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}$ where $\mathrm{g}\left(\mathrm{x}\right)$ and $\mathrm{f}\left(\mathrm{x}\right)$ are both differentiable functions then we substitute $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{u}$ which implies that ${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}=\mathrm{d}\mathrm{u}$.
• Therefore, the integral becomes $\int \mathrm{g}\left(\mathrm{u}\right)\mathrm{d}\mathrm{u}$ which can be solved by general formulas.

Solution:

In the given problem substitute $\ln \left(\mathrm{x}\right)=u$ therefore, $\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}=\mathrm{d}\mathrm{u}$.

The given integral becomes

$\int \cos \mathrm{u}\mathrm{d}\mathrm{u}=\sin \mathrm{u}+\mathrm{C}$

Resubstitute $\mathrm{u}=\ln \left(\mathrm{x}\right)$.

Concept:

$\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{x}+\mathrm{c}$

${\mathrm{x}}^{-1}=\frac{1}{\mathrm{x}}$

Calculation:

Let, I = $\int \frac{1}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}\mathrm{d}\mathrm{x}$

= $\int \frac{1}{{\mathrm{e}}^{\mathrm{x}}+\frac{1}{{\mathrm{e}}^{\mathrm{x}}}}\mathrm{d}\mathrm{x}$                (∵ ${\mathrm{x}}^{-1}=\frac{1}{\mathrm{x}}$)

= $\int \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{2\mathrm{x}}+1}\mathrm{d}\mathrm{x}$

Now, let e^x = t

⇒ ex dx = dt

∴ I = $\int \frac{\mathrm{d}\mathrm{t}}{{\mathrm{t}}^2+1}$

= $\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{t}+\mathrm{c}$             (∵ $\int \frac{1}{1+{\mathrm{x}}^2}\mathrm{d}\mathrm{x}=\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\mathrm{x}+\mathrm{c}$)

= $\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}({\mathrm{e}}^{\mathrm{x}})+\mathrm{c}$         (∵ ex = t)

Hence, option (4) is correct. 

Concept:

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\log |\mathrm{x}|+\mathrm{c}$

Calculation:

I = ∫ cot 2x dx

$=\int \frac{\cos 2\mathrm{x}}{\sin 2\mathrm{x}}\mathrm{d}\mathrm{x}$

Let sin 2x = t

Differentiating with respect to x, we get

⇒ 2 cos 2x dx = dt

⇒ cos 2x dx = $\frac{\mathrm{d}\mathrm{t}}{2}$

$\mathrm{I}=\frac{1}{2}\int \frac{1}{\mathrm{t}}\mathrm{d}\mathrm{t}$

= $\frac{1}{2}\log |\mathrm{t}|+\mathrm{c}$

= $\frac{1}{2}\log |\sin 2\mathrm{x}|+\mathrm{c}$

Concept:

Partial Fraction:

Calculation:

Here we have to find the value of $\int \frac{dx}{x\left(x+2\right)}$

Let $\frac{1}{x\left(x+2\right)}=\frac{A}{x}+\frac{B}{x+2}$

⇒ 1 = A (x + 2) + B x --------(1)

By putting x = 0 on both the sides of (1) we get A = 1/2

By putting x = - 2 on both the sides of (1) we get B = - 1/2

$\Rightarrow \frac{1}{x\left(x+2\right)}=\frac{1}{2x}-\frac{1}{2x+4}$

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

As we know that $\int \frac{dx}{x}=\log \left|x\right|+C$  where C is a constant

$\Rightarrow \int \frac{dx}{x\left(x+2\right)}=\frac{1}{2}\log \left|\frac{x}{x+2}\right|+C$where C is a constant

Concept:

sin²x + cos²x = 1

$\int \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}\mathrm{d}\mathrm{x}=-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}$

$\int \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}\mathrm{d}\mathrm{x}=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}$

Calculation:

We have $\int \sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{x}}\mathrm{d}\mathrm{x}$ = A sinx + B cosx + C

⇒ $\int (\sqrt{\mathrm{s}\mathrm{i}{\mathrm{n}}^2\mathrm{x}+\mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{x}-2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}})\mathrm{d}\mathrm{x}$ = A sinx + B cosx + C

⇒ $\int (\sqrt{(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}{)}^2})\mathrm{d}\mathrm{x}$ = A sinx + B cosx + C

⇒ $\int (|(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})|)\mathrm{d}\mathrm{x}$ = 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 

⇒ $\int (-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})\mathrm{d}\mathrm{x}$ = 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.

Formula Used:

(a³ - b³) = (a - b) × (a² + b² + ab)

∫ xⁿ = nx^n-1 + c

Calculation:

(x⁶ - 1) = (x² - 1) × (x⁴ + 1 + x²)

⇒ 5 × (x6 - 1)/(x2 - 1) = 5(x4 + 1 + x2)

Now,

I = $\int \frac{5(x^6-1)}{(x^2-1)}$ dx 

I = ∫ (5x4 + 5 + 5x²) dx

On integration 

⇒ 5x⁵/5 + 5x + 5x³/3 + c

⇒ x⁵ + 5x³/3 + 5x + c

∴ $x^5+\frac{5}{3}{x}^3+5x+c$