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New answer posted

7 months ago

Class 12th Maths Integrals

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 7 months ago

Class 12th Maths Integrals

Kindly consider the following

0 Follower 2 Views

alok kumar singh Contributor-Level 10

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

0 0

New answer posted 7 months ago

Class 12th Maths Integrals

Kindly consider the following

0 Follower 1 View

alok kumar singh Contributor-Level 10

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

0 0

New answer posted 7 months ago

Class 12th Maths Integrals

$\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t I = \int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x \\ P u t t a n^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} . d x = d t \\ = \int e^{t} (1 + t a n t + t a n^{2} t) d t = \int e^{t} (s e c^{2} t + t a n t) d t \\ H e r e, f (t) = t a n t \\ ∴ f^{'} (t) = s e c^{2} t \\ = e^{t} . f (t) = e^{t} t a n t = e^{t a n^{- 1} x} . x + C \\ [? \int e^{t} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C] \\ H e n c e, I = e^{t a n^{- 1} x} . x + C \end{array}$

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New answer posted

7 months ago

Class 12th Maths Integrals

$\int \frac{2 x - 1}{(x - 1) (x - 2) (x - 3)} d x$

0 Follower 1 View

alok kumar singh

Contributor-Level 10

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

0 0

New answer posted

7 months ago

Class 12th Maths Integrals

Kindly consider the following

0 Follower 1 View

alok kumar singh

Contributor-Level 10

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

0 0

New answer posted

7 months ago

Class 12th Maths Integrals

$\int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \\ P u t x^{2} = t f o r t h e p u r p o s e o f p a r t i a l f r a c t i o n . \\ W e g e t \frac{t}{(t + a^{2}) (t + b^{2})} \\ P u t \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A}{t + a^{2}} + \frac{B}{t + b^{2}} [w h e r e A a n d B a r e a r b i t r a r yconstants .] \\ \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A (t + b^{2}) + B (t + a^{2})}{(t + a^{2}) (t + b^{2})} \\ \Rightarrow t = A t + A b^{2} + B t + B a^{2} \\ C o m p a r i n g t h e l i k e t e r m s, w e g e t \\ A + B = 1 a n d A b^{2} + B a^{2} = 0 \\ \Rightarrow A = \frac{- a^{2}}{b^{2}} B \\ ∴ \frac{- a^{2}}{b^{2}} B + B = 1 \\ B (\frac{- a^{2}}{b^{2}} + 1) = 1 \Rightarrow B (\frac{- a^{2} + b^{2}}{b^{2}}) = 1 \\ \Rightarrow B = \frac{b^{2}}{b^{2} - a^{2}} a n d A = \frac{- a^{2}}{b^{2}} * \frac{b^{2}}{b^{2} - a^{2}} = \frac{a^{2}}{a^{2} - b^{2}} \\ S o, A = \frac{a^{2}}{a^{2} - b^{2}} a n d B = \frac{- b^{2}}{a^{2} - b^{2}} \\ ∴ \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = \frac{a^{2}}{a^{2} - b^{2}} \int \frac{1}{x^{2} + a^{2}} d x - \frac{b^{2}}{a^{2} - b^{2}} \int \frac{1}{x^{2} + b^{2}} d x \\ = \frac{a^{2}}{a^{2} - b^{2}} * \frac{1}{a} t a n^{- 1} \frac{x}{a} - \frac{b^{2}}{a^{2} - b^{2}} . \frac{1}{b} . t a n^{- 1} \frac{x}{b} \\ = \frac{a}{a^{2} - b^{2}} t a n^{- 1} \frac{x}{a} - \frac{b}{a^{2} - b^{2}} t a n^{- 1} \frac{x}{b} + C \\ H e n c e, I = \frac{1}{a^{2} - b^{2}} [a t a n^{- 1} \frac{x}{a} - b t a n^{- 1} \frac{x}{b}] + C . \end{array}$

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New answer posted

7 months ago

Class 12th Maths Integrals

Kindly consider the following

$\int^{} \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

0 0

New answer posted

7 months ago

Class 12th Chemical Kinetics

Consider the reaction A⇌B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?

0 Follower 15 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

If A→B then the concentration of both reactants and the products vary exponentially with time. But, in option B graph the reactant concentration decreases exponentially and the product concentration increases.

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New answer posted

7 months ago

Class 12th Chemical Kinetics

The value of rate constant of a pseudo first order reaction ____________.

A. Depends on the concentration of reactants present in small amount

B. Depends on the concentration of reactants present in excess.

C. Is independent of the concentration of reactants.

D. Depends only on temperature.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

Let's start with what a pseudo-first-order response is.

Although the pseudo-first-order reaction looks to be an order, it belongs to another order. It's a second-order reaction because it involves two reactants.

Let's have a look at a reaction.

CH₃Br + OH→CH₃OH + Br^-

So, the rate law for the reaction is

Rate = k [OH] [CH₃Br]

Rate = k [OH^- ] [CH₃Br] = k (constant) [CH₃Br] = K' [CH₃Br]

Only the concentration of CH₃Br will change during the reaction, and the rate will be determined by the reaction's modifications.

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