`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is ______.

`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is π. Explanation: Let `int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1)` dx `int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x) dx + int_((-pi)/2)^(pi/2) 1* dx` Because `(x^3 + x cos x + tan^5 x)` is an equivalent function. Hence, `int_(-pi//2)^(pi//2) (x^3 + x cos x + tan^5 x) dx = 0` `=> I = 0 + [x]_(-pi/2)^(pi/2)` `= pi/2 - (- pi/2)` `= pi/2 + pi/2 = pi`

∫-π2π2(x3+xcosx+tan5x+1)dx is ______. - Mathematics

प्रश्न

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + x \cos x + \tan^{5} x + 1) d x$ is ______.

विकल्प

MCQ

रिक्त स्थान भरें

उत्तर

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + x \cos x + \tan^{5} x + 1) d x$ is π.

Explanation:

Let $\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + x \cos x + \tan^{5} x + 1)$ dx

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + x \cos x + \tan^{5} x) d x + \int_{\frac{- π}{2}}^{\frac{π}{2}} 1 \cdot d x$

Because $(x^{3} + x \cos x + \tan^{5} x)$ is an equivalent function.

Hence, $\int_{- π / 2}^{π / 2} (x^{3} + x \cos x + \tan^{5} x) d x = 0$

$\Rightarrow I = 0 + {[x]}_{- \frac{π}{2}}^{\frac{π}{2}}$

$= \frac{π}{2} - (- \frac{π}{2})$

$= \frac{π}{2} + \frac{π}{2} = π$

shaalaa.com

Properties of Definite Integrals

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 20 Q 19 Q 21

अध्याय 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 12

अध्याय 7 Integrals
Exercise 7.11 | Q 20 | पृष्ठ ३४७

वीडियो ट्यूटोरियलVIEW ALL [5]

view
वीडियो ट्यूटोरियल For All Subjects
Properties of Definite Integrals

video tutorial00:07:37
Properties of Definite Integrals

video tutorial01:56:53
Properties of Definite Integrals

video tutorial00:26:24
Properties of Definite Integrals

video tutorial00:41:32
Properties of Definite Integrals

video tutorial00:51:59

संबंधित प्रश्न

Prove that: $\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$

Evaluate : $\int_{0}^{π} \frac{x \sin x}{1 + \sin x} d x$

If $\int_{0}^{α} (3 x^{2} + 2 x + 1) d x = 14$ then $α =$

(A) 1

(B) 2

(D) –2

By using the properties of the definite integral, evaluate the integral:

$\int_{0}^{\frac{π}{2}} \cos^{2} x d x$

By using the properties of the definite integral, evaluate the integral:

$\int_{0}^{\frac{π}{2}} (2 \log \sin x - \log \sin 2 x) d x$

By using the properties of the definite integral, evaluate the integral:

$\int_{0}^{π} \log (1 + \cos x) d x$

By using the properties of the definite integral, evaluate the integral:

$\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x$

The value of $\int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x})$ dx is ______.

$\int_{0}^{a} 3 x^{2} d x = 8,$ find the value of a.

If $f (a + b - x) = f (x)$ , then prove that

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

Choose the correct alternative:

$\int_{- 9}^{9} \frac{x^{3}}{4 - x^{2}} d x$ =

By completing the following activity, Evaluate $\int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{7 - x}} d x$ .

Solution: Let I = $\int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{7 - x}} d x$ ......(i)

Using the property, $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ , we get

I = $\int_{2}^{5} \frac{( )}{\sqrt{7 - x} + ( )} d x$ ......(ii)

Adding equations (i) and (ii), we get

2I = $\int_{2}^{5} \frac{\sqrt{x}}{\sqrt{x} - \sqrt{7 - x}} d x + () d x$

2I = $\int_{2}^{5} (\frac{( ) + ( )}{( ) + ( )}) d x$

2I = $□$

∴ I = $□$

$\int \frac{\cos x + x \sin x}{x (x + \cos x)}$ dx = ?

$\int_{0}^{\frac{π}{4}} \frac{\sec^{2} x}{(1 + \tan x) (2 + \tan x)}$ dx = ?

The c.d.f, F(x) associated with p.d.f. f(x) = 3(1- 2x²). If 0 < x < 1 is k $(x - \frac{2 x^{3}}{k})$ , then value of k is ______.

$\int_{0}^{1} (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + ... upto \infty)$ e^2x dx = ?

$\int_{0}^{\frac{π}{4}} \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x$ = ____________

$\int_{0}^{1} x \tan^{- 1} x d x$ = ______

$\int_{\frac{π}{6}}^{\frac{π}{3}} \sin^{2} x d x$ = ______

$\int_{0}^{π} \sin^{2} x . \cos^{2} x d x$ = ______

$\int_{- 1}^{1} \log (\frac{2 - x}{2 + x}) dx = ?$

$\int_{0}^{9} \frac{1}{1 + \sqrt{x}}$ dx = ______

Evaluate $\int_{0}^{\frac{π}{2}} \frac{\tan^{7} x}{\cot^{7} x + \tan^{7} x} d x$

Find $\int_{0}^{\frac{π}{4}} \sqrt{1 + \sin 2 x} d x$

Show that $\int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\sin x + \cos x} = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Evaluate the following:

$\int_{0}^{\frac{π}{2}} \frac{dx}{{(a^{2} \cos^{2} x + b^{2} \sin^{2} x)}^{2}}$ (Hint: Divide Numerator and Denominator by cos⁴x)

Evaluate the following:

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \log | \sin x + \cos x | d x$

$\int_{0}^{\frac{π}{2}} \cos x e^{\sin x} d x$ is equal to ______.

$\int \frac{d x}{e^{x} + e^{- x}}$ is equal to ______.

Evaluate: $\int_{1}^{3} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4} - x} d x$

If $\int_{0}^{\frac{π}{2}} \log \cos x d x = \frac{π}{2} \log (\frac{1}{2})$ , then $\int_{0}^{\frac{π}{2}} \log \sec d x$ = ______.

$\int_{1}^{2} x \log x d x$ = ______

$\int_{0}^{2 a} \frac{f (x)}{f (x) + f (2 a - x)} d x$ = ______

Evaluate the following integral:

$\int_{0}^{1} x {(1 - x)}^{5} d x$

Evaluate the following integrals:

$\int_{- 9}^{9} \frac{x^{3}}{4 - x^{3}} d x$

Solve.

$\int_{0}^{1} e^{x^{2}} x^{3} d x$

Evaluate:

$\int_{0}^{6} | x + 3 | d x$

Evaluate the following integral:

$\int_{0}^{1} x {(1 - x)}^{5} d x$

Evaluate the following definite integral:

$\int_{- 2}^{3} \frac{1}{x + 5} d x$

∫-π2π2(x3+xcosx+tan5x+1)dx is ______. - Mathematics

Advertisements

Advertisements

प्रश्न

विकल्प

उत्तर

APPEARS IN

वीडियो ट्यूटोरियलVIEW ALL [5]

संबंधित प्रश्न

Select a course

∫-π2π2(x3+xcosx+tan5x+1)dx is ______. - Mathematics

Advertisements

Advertisements

प्रश्न

विकल्प

उत्तरShow Solution

APPEARS IN

वीडियो ट्यूटोरियलVIEW ALL [5]

संबंधित प्रश्न

Select a course

उत्तर