## $\int \frac{(x + x^{2/3} + x^{1/6})}{x(1 + x^{1/3})} dx$ is equal to

Q: $\large \int \frac{(x + x^{2/3} + x^{1/6})}{x(1 + x^{1/3})} dx$ is equal to

(A) $\large \frac{3}{2}x^{2/3} + 6 tan^{-1}(x^{1/6}) + c$

(B) $\large \frac{3}{2}x^{2/3} – 6 tan^{-1}(x^{1/6}) + c$

(C) $\large \frac{3}{2}x^{2/3} + tan^{-1}(x^{1/6}) + c$

(D) none of these

Sol:  Substituting x = z6, dx = 6 z5dz , we have

$\large I = \int \frac{6 z^5 (z^6 + z^4 + z)}{z^6 (1+z^2)}dz$

$\large = \int \frac{6 (z^5 + z^3 + 1)}{(1+z^2)}dz$

$\large = \int 6 z^3 dz + \int (\frac{6 dz}{z^2 + 1})$

$\large = 6\frac{z^4}{4} + 6 tan^{-1}z + c$

$\large \frac{3}{2}x^{2/3} + 6 tan^{-1}(x^{1/6}) + c$

Hence (A) is the correct answer.

## $\int \frac{x^2 -1}{(x^2 +1)\sqrt{x^4 +1}} dx$ is equal to

Q: $\large \int \frac{x^2 -1}{(x^2 +1)\sqrt{x^4 +1}} dx$ is equal to

(A) $\large sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c$

(B) $\large \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c$

(C) $\large \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}}) + c$

(D) none of these

Sol: $\large I = \int \frac{x^2(1 -\frac{1}{x^2})}{x^2(x +\frac{1}{x})\sqrt{x^2 + \frac{1}{x^2}}} dx$

Let $\large x + \frac{1}{x} = z$ ; $\large (1-\frac{1}{x^2})dx = dz$

$\large I = \int \frac{dz}{z \sqrt{z^2 -2}}$

$\large = \frac{1}{\sqrt{2}} sec^{-1}\frac{z}{\sqrt{2}}$

$\large = \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c$

## If f'(x) = x |sinx| ∀ x  ∈ (0, π) and f(0) = 1/√3 , then f(x) will be

Q: If f'(x) = x |sinx| ∀ x  ∈ (0, π) and f(0) = 1/√3 , then f(x) will be

(A) – x cosx + sinx + 1/√3

(B) 1/√3 – x cosx + sinx

(C) sinx + x cosx – 2/√3

(D) sinx + x cosx + 2/√3

Sol. f'(x) = x |sinx| ∀ x  ∈ (0, π)

$\large \int f'(x) dx = \int x sinx dx$

$\large f(x) = – x cosx + \int cosx dx + c$

$\large = – x cosx + sinx + c$

f(0) = c = 1/√3

$\large f(x) = – x cosx + sinx + \frac{1}{\sqrt{3}}$

Hence (A) is the correct answer.

## $\int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx$ is equal to

Q: $\large \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx$ is equal to

(A) $\large \frac{1}{2} ln|sin^2 x + sin^2 2x| + c$

(B) $\large \frac{1}{2}ln |sinx| + \frac{1}{2} ln|sin x + 4sinx cos^2x| + c$

(C) $\large \frac{1}{2} ln|sin^2 x – sin^2 2x| + c$

(D) $\large \frac{1}{2}ln |sinx| – ln|sin x + 4sinx cos^2x| + c$

Sol: $\large \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx$

Put cosx = t

$\large = – \int \frac{t (1 + 4(2t^2 – 1))}{1-t^2 + 4 (1-t^2)t^2} dt$

$\large = \int \frac{t(8t^2 – 3)}{4t^4 – 3t^2 – 1} dt$

$\large = \frac{1}{2} ln |4t^4 – 3t^2 – 1| + c$

$\large = \frac{1}{2}ln |sinx| + \frac{1}{2} ln|sin x + 4sinx cos^2x| + c$

## $\int ( x + \frac{1}{x})^{n+5} (\frac{x^2 -1}{x^2})dx$ is equal to

Q: $\int ( x + \frac{1}{x})^{n+5} (\frac{x^2 -1}{x^2})dx$ is equal to

(A) $\frac{(x+ \frac{1}{x})^{n+6}}{n+6} + c$

(B) $(\frac{x^2 + 1}{x^2})^{n+6} (n+6) + c$

(C) $(\frac{x}{x^2 + 1})^{n+6} (n+6) + c$

(D) none of these

Sol: Let $(x+ \frac{1}{x}) = z$ then $(1 – \frac{1}{x^2})dx = dz$

$I = \int z^{n+5} dz = \frac{z^{n+6}}{n+6} + c$

$= \frac{(x + \frac{1}{x})^{n+6}}{n+6} + c$

Hence (A) is the correct answer.

## If $\large \int e^{ax} cos bx dx = \frac{e^{2x}}{2a}f(x) + c$ , then

Q: If $\large \int e^{ax} cos bx dx = \frac{e^{2x}}{2a}f(x) + c$ , then

(A) $\large \frac{f”(x)}{f(x)} = 24$

(B) $\large \frac{f”(x)}{f(x)} = 25$

(C) range of f(x) is $(-\sqrt{29} , \sqrt{29})$

(D) $\large \frac{f(x)}{f'(x)} = e^{cosbx} (a sinx)$

Sol: $\large \int e^{ax} cos bx dx = \frac{e^{ax}}{a^2 + b^2}(a cosbx + b sinbx) + c$

$\large \frac{e^{2x}}{2^2 + 5^2}f(x) + c = \frac{e^{ax}}{a^2 + b^2}(a cosbx + b sinbx) + c$

$\large \frac{e^{2x}}{29}f(x) + c = \frac{e^{ax}}{a^2 + b^2}(a cosbx + b sinbx) + c$

a = 2 , b = 5 and f(x) = 2 cos5x + 5 sin5x

Range of f(x) is $(-\sqrt{29} , \sqrt{29})$

Hence (C) is the correct answer.

## If $I_{m,n} = \int cos^m x cosnx dx$ then the value of (m + n)Im,n – m Im–1,n–1 (m , n ∈ N) is equal to

Q: If $I_{m,n} = \int cos^m x cos nx dx$ then the value of (m + n)Im,n – m Im–1,n–1 (m , n ∈ N) is equal to

(A) $\large \frac{cos^m x sin nx}{n} + c$

(B) $\large cos^m x sin nx + c$

(C) $\large \frac{cos^m x cos nx}{n} + c$

(D) $\large – cos^m x cos nx + c$

Sol: $\large I_{m,n} = \int cos^m x cos nx dx$

$\large = \frac{cos^m x sin nx}{n} + \frac{m}{n} \int cos^{m-1} x sinx sin nx dx$

$\large = \frac{cos^m x sin nx}{n} + \frac{m}{n} \int cos^{m-1}x ( cos(n-1)x – cosnx cosx ) dx$

$\large I_{m,n} = \frac{cos^m x sin nx}{n} + \frac{m}{n} \int cos^{m-1}x cos(n-1)x dx – \frac{m}{n}I_{m,n}$

$\large \frac{m+n}{n} I_{m,n} = \frac{cos^m x sin nx}{n} + \frac{m}{n}I_{m-1,n-1} + c_1$

$\large (m+n)I_{m,n} = m I_{m-1,n-1} + cos^m x sin nx + c$

Hence (B) is the correct answer.

## $\int \frac{x^2 -1}{x^3 \sqrt{2x^4 – 2x^2 +1 }} dx$ is equal to

Q: $\large \int \frac{x^2 -1}{x^3 \sqrt{2x^4 – 2x^2 +1 }} dx$ is equal to

(A) $\large \frac{\sqrt{2x^4 – 2x^2 +1}}{x^2} + c$

(B) $\large \frac{\sqrt{2x^4 – 2x^2 +1}}{x^3} + c$

(C) $\large \frac{\sqrt{2x^4 – 2x^2 +1}}{x} + c$

(D) $\large \frac{\sqrt{2x^4 – 2x^2 +1}}{2x^2} + c$

Sol: $\large \int \frac{\frac{1}{x^3} – \frac{1}{x^5}}{\sqrt{2 – \frac{2}{x^2} + \frac{1}{x^4}}} dx$

Let $\large 2 – \frac{2}{x^2} + \frac{1}{x^4} = z$

$\large = \frac{1}{4} \int \frac{dz}{\sqrt{z}}$

$\large = \frac{1}{2} \sqrt{z} + c$

$\large = \frac{1}{2} \sqrt{2 – \frac{2}{x^2} + \frac{1}{x^4}} + c$

Hence (D) is the correct answer.