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Integrals Class 12th

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

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New answer posted

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Integrals Class 12th

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x \end{array}$

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New answer posted 3 months ago

Integrals Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Kindly consider the following $\int \frac{d x}{1 + c o s x}$

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int \frac{d x}{1 + c o s x} \\ = \int \frac{d x}{2 c o s^{2} x / 2} [? 1 + c o s x = 2 c o s^{2} x / 2] \\ = \frac{1}{2} \int s e c^{2} \frac{x}{2} d x = \frac{1}{2} . 2 t a n \frac{x}{2} + C = t a n \frac{x}{2} + C \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s t a n \frac{x}{2} + C . \end{array}$

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3 months ago

Integrals Maths Integrals

271. The value of $\int_{0}^{1} t a n^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ is

A.1

B. 0

C.-1

D. $\frac{π}{4}$

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Vishal Baghel

Contributor-Level 10

Consider, $I = \int_{0}^{1} t a n^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$

$\begin{array}{l} \Rightarrow I = \int_{0}^{1} t a n^{- 1} (\frac{x - (1 - x)}{1 + x (1 - x)}) d x \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} x - t a n^{- 1} (1 - x)] d x - - - - - (1) \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} (1 - x) - t a n^{- 1} (1 - 1 + x)] d x \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} (1 - x) - t a n^{- 1} x] d x \\ \Rightarrow I = \int_{0}^{1} [t a n^{- 1} (1 - x) - t a n^{- 1} (x)] d x - - - - - (2) \end{array}$

Adding (1) and (2), we get

$\begin{array}{l} \Rightarrow 2 I = \int_{0}^{1} [t a n^{- 1} x - t a n^{- 1} (1 - x) - t a n^{- 1} (1 - x) - t a n^{- 1} x] d x \\ \Rightarrow 2 I = 0 \\ \Rightarrow I = 0 \end{array}$

Thus, the correct option is B.

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Integrals Class 12th

270. If $f (a + b - x) = f (x), t h e n, \int_{a}^{b} x f (x)$ is Equal to

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Vishal Baghel

Contributor-Level 10

Giver, f(a + bx) = f(x). _________ (1)

Let I $= \int_{a}^{b} x + (x) = \int_{a}^{b} (a + b - x) f (a + b - x) d x_____(2)$

${? \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x .$

$= \int_{a}^{b} (a + b - x) f (x) d x .$ ___________ (3) {because Equation (1)}

$+ = \int_{a}^{b} [(a + b - x) f (x) + x f (x)] d x$

$\Rightarrow 2I = \int_{a}^{b} (a + b - x + x) \overset{}{f} (x) d x$

$\Rightarrow 2I = \int_{a}^{b} (a + b) f (x) d x$

$\Rightarrow = \frac{a + b}{2} \int_{a}^{b} f (x) d x .$

therefore, Option D is correct.

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Integrals Maths Integrals

269. $\int \frac{c o s 2 x d x}{{(s i n x + c o s x)}^{2}}$ is Equal to

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Vishal Baghel

Contributor-Level 10

Let $= \int \frac{c o s 2 x d x}{{(s i n x + c o s x)}^{2}}$

$= \int \frac{c o s^{2} x - s i n^{2} x}{{(s i n x + c o s x)}^{2}} d x {? c o s 2 x = c o s^{2} x - s i n^{2} x$

$= \int \frac{(c o s x + s i n x) (c o s x - s i n x)}{{(s i n x + c o s x)}^{2}} d x {? a^{2} - b^{2} = (x + b) (x - b)$

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

$= l o g | s i n x + c o s x | + c {\int \frac{f^{'} (x)}{f (x)} d x = l o g | f (x) | + c$

So, option B is correct.

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Integrals Maths Integrals

268. $\int \frac{d x}{e^{x} + e^{- x}}$ is Equal to

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Vishal Baghel

Contributor-Level 10

Let $= \int \frac{d x}{e^{x} + e^{- x}}$

$= \int \frac{d x}{e^{x} + \frac{1}{e^{x} .}} = \int \frac{e^{x} d x}{e^{x} \cdot e^{x} + 1}$

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

Putting e^x = t

e^xdx = dt.

$∴ = \int \frac{d t}{t^{2} + 1 .}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

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Integrals Maths Integrals

267. Evaluate $\int_{0}^{1} e^{2 - 3 x} d x$ as a limit of a sum.

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Vishal Baghel

Contributor-Level 10

Let $I = \int_{0}^{1} e^{2 - 3 x} d x$

We know that,

$\int_{a}^{b} f (x) d x = (b - a) \underset{n \to \infty}{l i m} \frac{1}{n} [f (a) + f (a + h) + . . . + f (a (n - 1) h)]$

Where, $h = \frac{b - a}{n}$

Here, $a = 0, b = 1$ and $f (x) e^{2 - 3 x}$

$\Rightarrow h = \frac{1 - 0}{n} = \frac{1}{n}$

$∴ \int_{0}^{1} e^{2 - 3 x} d x = (1 - 0) \underset{n \to \infty}{l i m} \frac{1}{n} [f (0) + f (0 + h) + . . . + f (0 + (n - 1) h)]$

$\begin{array}{l} = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} + e^{2 - 3 x} + . . . + e^{2 - 3 (n - 1) h}] = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} {1 + e^{- 3 h} + e^{- 6 h} + e^{- 9 h} + . . . + e^{- 3 (n - 1) h}}] \\ = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} {\frac{1 - {(e^{- 3 h})}^{n}}{1 - (e^{- 3 h})}}] = \underset{n \to \infty}{l i m} \frac{1}{n} [e^{2} {\frac{1 - e^{\frac{- 3}{n} n}}{1 - e^{\frac{- 3}{n}}}}] \\ = \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{e^{2} (1 - e^{- 3})}{1 - e^{\frac{- 3}{n}}}] = e^{2} (e^{- 3} - 1) \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{1}{e^{\frac{- 3}{n}} - 1}] \\ = \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{e^{2} (1 - e^{- 3})}{1 - e^{\frac{- 3}{n}}}] = e^{2} (e^{- 3} - 1) \underset{n \to \infty}{l i m} \frac{1}{n} [\frac{1}{e^{\frac{- 3}{n}} - 1}] \end{array}$

$\begin{array}{l} = e^{2} (e^{- 3} - 1) \underset{n \to \infty}{l i m} (- \frac{1}{3}) [\frac{- \frac{3}{n}}{e^{\frac{- 3}{n}} - 1}] = \frac{e^{2} (e^{- 3} - 1)}{3} \underset{n \to \infty}{l i m} [\frac{- \frac{3}{n}}{e^{\frac{- 3}{n}} - 1}] \\ = \frac{- e^{2} (e^{- 3} - 1)}{3} (1) \\ = \frac{- e^{- 1} + e^{2}}{3} \\ = \frac{1}{3} (e^{2} - \frac{1}{e}) \end{array}$

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Integrals Maths Integrals

266. $\int_{0}^{1} s i n^{- 1} x d x . = \frac{π}{2} - 1$

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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New answer posted

3 months ago

Integrals Maths Integrals

265. $\int_{0}^{\frac{π}{4}} \cdot 2 t a n^{3} x d x = 1 - l o g 2$

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Vishal Baghel

Contributor-Level 10

Let I $\int_{0}^{\frac{π}{4}} \cdot 2 t a n^{3} x d x$

$= 2 \int_{0}^{\frac{π}{4}} t a n^{2} x t a n x d x$

$= 2 \int_{0}^{\frac{π}{4}} (s e c^{2} x - 1) t a n x d x {? s e c^{2} x = t a n^{2} x + 1}$

$= 2 \int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x - 2 \int_{0}^{\frac{π}{4}} t a n x d x$

$= 2I_{1} + 2 [l o g] {(c o s x)}_{0}^{\frac{π}{4}}$

$= 2I_{1} + 2 [l o g (c o s \frac{π}{4}) - l o g (c o s 0)]$

$= 2I_{1} + 2 (l o g1/√2$
$\frac{}{} - l o g 1)$

$= 2I_{1} + 2 (l o g \cdot 2^{\frac{- 1}{2}} - 0)$

$I = 2I_{1} + 2 * (\frac{- 1}{2}) l o g 2 = 2I_{1} - l o g 2 .____(1) .$

Where I $_{1} = \int_{0}^{\frac{π}{4}} s e c^{2} x t a n x d x$

Let tan x = t =>sec²xdx = dt

When, x = 0, t = tan 0. = 0

$x = \frac{π}{4}, t = t a n \frac{π}{4} = 1$

$_{1} = \int_{0}^{1} t d t = {[\frac{t^{2}}{2}]}_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}$

So, Equation (1) becomes,

$= 2 * \frac{1}{2} - l o g 2$

=1 - log 2.

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