Indefinite Integration(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

INDEFINITE INTEGRATION MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

If $I_{n} = \int t a n^{n} x d x, t h e n I_{0} + I_{1} + 2 (I_{2} + . . . I_{8}) + I_{9} + I_{10}$ , is equal to

$\sum_{n = 1}^{9} \frac{t a n^{n} x}{n}$
$1 + \sum_{n = 1}^{8} \frac{t a n^{n} x}{n}$
$\sum_{n = 1}^{9} \frac{t a n^{n} x}{n + 1}$
$\sum_{n = 2}^{10} \frac{t a n^{n} x}{n + 1}$

Discuss Question

Answer: Option A. ->

\sum_{n = 1}^{9} \frac{t a n^{n} x}{n}

:
A
We have

I_{n} = \int t a n^{n} x d x = \int t a n^{n - 2} x (s e c^{2} x - 1) d x = \frac{t a n^{n - 1} x}{n - 1} - I_{n - 2} i . e . I_{n - 2} + I_{n} = \frac{t a n^{n - 1} x}{n - 1} (n \geq 2)

Thus, we have

I_{0} + I_{1} + 2 (I_{2} + . . . . + I_{8}) + I_{9} + I_{10} = (I_{0} + I_{2}) + (I_{1} + I_{3}) + (I_{2} + I_{4}) + (I_{3} + I_{5}) + (I_{4} + I_{6}) + (I_{5} + I_{7}) + (I_{6} + I_{8}) + (I_{7} + I_{9}) + (I_{8} + I_{10}) . = \sum_{n = 2}^{10} \frac{t a n^{n - 1} x}{n - 1} = \sum_{n = 1}^{9} \frac{t a n^{n} x}{n}

Question 2.

Primitive of $f (x) = x {.2}^{I n (x^{2} + 1)}$ with respect to x is

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

Discuss Question

Answer: Option C. ->

\frac{(x^{2} + 1)^{I n 2 + 1}}{2 (I n 2 + 1)} + C

:
C

I = \int x 2^{I n (x^{2} + 1)} d x

let

x^{2} + 1 = t; x d x = \frac{d t}{2}

Hence

I = \frac{1}{2} \int 2^{I n t} d t = \frac{1}{2} \int t^{I n 2} d t = \frac{1}{2} . \frac{t^{I n 2 + 1}}{I n 2 + 1} + C = \frac{1}{2} . \frac{(x^{2} + 1)^{I n 2 + 1}}{I n 2 + 1} + C

Question 3.

$\int \frac{c o s^{3} x + c o s^{5} x}{s i n^{2} x + s i n^{4} x} d x$ equals

$s i n x - 6 t a n^{- 1} (s i n x) + c$
$s i n x - 2 s i n^{- 1} x + c$
$s i n x - 2 (s i n x)^{- 1} - 6 t a n^{- 1} (s i n x) + c$
$s i n x - 2 (s i n x)^{- 1} + 5 t a n^{- 1} (s i n x) + c$

Discuss Question

Answer: Option C. ->

s i n x - 2 (s i n x)^{- 1} - 6 t a n^{- 1} (s i n x) + c

:
C

s i n x = t; I = \int \frac{(1 - t^{2}) (2 - t^{2})}{t^{2} (1 + t^{2})} d t = \int (1 + \frac{2}{t^{2}} - \frac{6}{1 + t^{2}}) d t

= s i n x - 2 (s i n x)^{- 1} - 6 t a n^{- 1} (s i n x) + c

Question 4.

$\int \frac{d x}{(x - 3)^{(4 / 5)} (x + 1)^{6 / 5}} =$

$((x - 3) (x + 1)^{(1 / 5)}) + c$
$\frac{5}{4} {(\frac{x - 3}{x + 1})}^{1 / 5} + c$
${(\frac{x - 3}{x + 1})}^{1 / 5} + c$
$(x - 3)^{6 / 5} (x + 1)^{4 / 5} + c$

Discuss Question

Answer: Option B. ->

\frac{5}{4} {(\frac{x - 3}{x + 1})}^{1 / 5} + c

:
B

I = \int \frac{d x}{(x - 3)^{4 / 5} (x + 1)^{6 / 5}} P u t t = \frac{x - 3}{x + 1} d t = \frac{4}{(x + 1)^{2}} d x . I = \frac{1}{4} \int \frac{d t}{t^{4 / 5}} = \frac{1}{4} (\frac{t^{- 4 / 5} + 1}{- 4 / 5} + 1) = \frac{5}{4} t^{1 / 5} = \frac{5}{4} {(\frac{x - 3}{x + 1})}^{1 / 5}

Question 5.

Let $x^{2} \neq n π - 1, n ϵ N$ . Then, the value of $\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x$ is equal to

$l o g | \frac{1}{2} s e c (x^{2} + 1) | + C$
$l o g | s e c (\frac{x^{2} + 1}{2}) | + C$
$\frac{1}{2} l o g | s e c (x^{2} + 1) | + C$
None of these

Discuss Question

Answer: Option B. ->

l o g | s e c (\frac{x^{2} + 1}{2}) | + C

:
B
We have,

\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x = \int x \sqrt{\frac{2 s i n (x^{2} + 1) - 2 s i n (x^{2} + 1) c o s (x^{2} + 1)}{2 s i n (x^{2} + 1) + 2 s i n (x^{2} + 1) c o s (x^{2} + 1)}} d x = \int x \sqrt{\frac{1 - c o s (x^{2} + 1)}{1 + c o s (x^{2} + 1)}} d x = \int x t a n (\frac{x^{2} + 1}{2}) d x = \int t a n (\frac{x^{2} + 1}{2}) d (\frac{x^{2} + 1}{2}) = l o g | s e c (\frac{x^{2} + 1}{2}) | + C

Question 6.

$\int \frac{d x}{c o s (2 x) c o s (4 x)}$ is equal to

$\frac{1}{2 \sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 - \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |) + C$
$\frac{1}{2 \sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 + \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |) + C$
$\frac{1}{\sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 + \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x - t a n 2 x |) + C$
None of these

Discuss Question

Answer: Option A. ->

\frac{1}{2 \sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 - \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |) + C

:
A

\int \frac{s i n (4 x - 2 x) d x}{s i n (2 x) c o s (2 x) c o s (4 x)} = \int \frac{s i n (4 x) d x}{s i n (2 x) c o s (4 x)} - \int s e c 2 x d x = 2 \int \frac{c o s 2 x d x}{c o s 4 x} - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |)

Question 7.

$\int \frac{1 - 7 c o s^{2} x}{s i n^{7} x c o s^{2} x} d x = \frac{f (x)}{(s i n x)^{7}} + C$ , then f(x) is equal to

sin x
cos x
tan x
cot x

Discuss Question

Answer: Option C. -> tan x
:
C

\int \frac{1 - 7 c o s^{2} x}{s i n^{7} x c o s^{2} x} d x = \int (\frac{s e c^{2} x}{s i n^{7} x} - \frac{7}{s i n^{7} x}) d x = \int \frac{s e c^{2} x}{s i n^{7} x} d x - \int \frac{7}{s i n^{7}} d x = I_{1} + I_{2} N o w, I_{1} = \int \frac{s e c^{2} x}{s i n^{7}} d x = \frac{t a n x}{s i n^{7} x} + 7 \int \frac{t a n x c o s x}{s i n^{8} x} = \frac{t a n x}{s i n^{7} x} - I_{2} ∴ I_{1} + I_{2} = \frac{t a n x}{s i n^{7} x} + C \Rightarrow f (x) = t a n x

Question 8.

$\int e^{x} [\frac{x^{3} + x + 1}{(1 + x^{2})^{3 / 2}}] d x$ is equal to

$\frac{x^{2} e^{x}}{(1 + x^{2})^{1 / 2}} + c$
$\frac{e^{x}}{x (1 + x^{2})^{1 / 2}} + c$
$\frac{e^{x}}{(1 + x^{2})^{1 / 2}} + c$
$\frac{x e^{x}}{(1 + x^{2})^{1 / 2}} + c$

Discuss Question

Answer: Option D. ->

\frac{x e^{x}}{(1 + x^{2})^{1 / 2}} + c

:
D

$I = \int e^{x} [\frac{x}{\sqrt{1 + x^{2}}} + \frac{1}{(1 + x^{2})^{3 / 2}}] d x$
Let $f (x) = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow f^{^{'}} (x) = \frac{\sqrt{1 + x^{2}} - \frac{x^{2}}{\sqrt{1 + x^{2}}}}{(1 + x^{2})}$
$= \frac{1}{(1 + x^{2})^{3 / 2}}$
$∴ I = e^{x} f (x) + c$
$= \frac{e^{x} x}{\sqrt{1 + x^{2}}} + c$

Question 9.

$\int \frac{dx}{{sin}^{4} x + {cos}^{4} x}$ is equal to

$\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C$
$\sqrt{2} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C$
$\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} cot 2 x) + C$
None of these

Discuss Question

Answer: Option A. ->

\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C

:
A

\int \frac{dx}{{sin}^{4} x + {cos}^{4} x} = \int \frac{dx}{({sin}^{2} x + {cos}^{2} x)^{2} - 2 {sin}^{2} x {cos}^{2} x} = \int \frac{2 dx}{2 - {sin}^{2} 2 x} = \int \frac{2 {sec}^{2} 2 x}{2 {sec}^{2} 2 x - {tan}^{2} 2 x} dx = 2 \int \frac{{sec}^{2} 2 x}{2 + {tan}^{2} 2 x} dx = \int \frac{dt}{(\sqrt{2})^{2} + t^{2}} [p u t t i n g tan 2 x = t \Rightarrow 2 {sec}^{2} 2 x dx = dt] = \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{t}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C

Question 10.

If $\int \frac{d x}{x \sqrt{1 - x^{3}}} = a l o g ∣ ∣ ∣ \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} ∣ ∣ ∣ + C$ then a =

$\frac{1}{3}$
$\frac{2}{3}$
$- \frac{1}{3}$
$- \frac{2}{3}$

Discuss Question

Answer: Option A. ->

\frac{1}{3}

:
A

Put<br> $1 - x^{3} = t^{2} \Rightarrow - 3 x^{2} d x = 2 t d t ∴ \int \frac{d x}{x \sqrt{1 - x^{3}}} = \int \frac{x^{2}}{x^{3} \sqrt{1 - x^{3}}} d x = \frac{2}{3} \int \frac{d t}{t^{2} - 1} = \frac{1}{3} l o g ∣ ∣ \frac{t - 1}{t + 1} ∣ ∣ + C = \frac{1}{3} l o g ∣ ∣ ∣ \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} ∣ ∣ ∣ + C ∴ a = \frac{1}{3}$