Question and Answers Forum

reduction formulas for n∈N, some n>0, some n>1 ∫sin^n x dx=−(1/n)cos x sin^(n−1) x +((n−1)/n)∫sin^(n−2) x dx ∫cos^n x dx=(1/n)sin x cos^(n−1) x +((n−1)/n)∫cos^(n−2) x dx ∫tan^n x dx=(1/(n−1))tan^(n−1) x −∫tan^(n−2) x dx ∫sec^n x dx=(1/(n−1))tan x sec^(n−2) x +((n−2)/(n−1))∫sec^(n−2) x dx ∫csc^n x dx=−(1/(n−1))cot x csc^(n−2) x +((n−2)/(n−1))∫csc^(n−2) x dx ∫cot^n x dx=−(1/(n−1))cot^(n−1) x −∫cot^(n−2) x dx

$reduction formulas for n \in N, some n > 0, some n > 1$ $\int \sin^{n} x d x = - \frac{1}{n} \cos x \sin^{n - 1} x + \frac{n - 1}{n} \int \sin^{n - 2} x d x$ $\int \cos^{n} x d x = \frac{1}{n} \sin x \cos^{n - 1} x + \frac{n - 1}{n} \int \cos^{n - 2} x d x$ $\int \tan^{n} x d x = \frac{1}{n - 1} \tan^{n - 1} x - \int \tan^{n - 2} x d x$ $\int \sec^{n} x d x = \frac{1}{n - 1} \tan x \sec^{n - 2} x + \frac{n - 2}{n - 1} \int \sec^{n - 2} x d x$ $\int \csc^{n} x d x = - \frac{1}{n - 1} \cot x \csc^{n - 2} x + \frac{n - 2}{n - 1} \int \csc^{n - 2} x d x$ $\int \cot^{n} x d x = - \frac{1}{n - 1} \cot^{n - 1} x - \int \cot^{n - 2} x d x$

Question Number 63976 Answers: 0 Comments: 3

∫secxdx ?

$\int s e c x d x ?$

Question Number 63927 Answers: 0 Comments: 7

∫_0 ^π (dx/((3+2cos x)^2 ))

$\int_{0}^{π} \frac{d x}{{(3 + 2 c o s x)}^{2}}$

Question Number 63903 Answers: 0 Comments: 0

Question Number 63892 Answers: 0 Comments: 3

calculate A=∫_0 ^∞ (x^(2017) /(1+x^(2019) )) dx and B =∫_0 ^∞ (x^(2019) /(1+x^(2021) )) dx calculate the fraction (A/B)

$c a l c u l a t e A = \int_{0}^{\infty} \frac{x^{2017}}{1 + x^{2019}} d x a n d B = \int_{0}^{\infty} \frac{x^{2019}}{1 + x^{2021}} d x$ $c a l c u l a t e t h e f r a c t i o n \frac{A}{B}$

Question Number 63881 Answers: 0 Comments: 1

∫e^x /Lnxdx

$\int e^{x} / L n x d x$

Question Number 63883 Answers: 0 Comments: 1

∫ln(x)ln(1−x)ln(1−2x)dx

$\int l n (x) l n (1 - x) l n (1 - 2 x) d x$

Question Number 63852 Answers: 0 Comments: 0

prove that ∫_0 ^1 arctan(x) cot(((πx)/2)) dx = ((3 ln^2 (2))/(2π))+((lnπ ln2)/π)+∫_0 ^∞ ((ln(1+x^2 ))/(e^(2πx) +1)) dx

$p r o v e t h a t$ $\int_{0}^{1} a r c t a n (x) c o t (\frac{π x}{2}) d x = \frac{3 {l n}^{2} (2)}{2 π} + \frac{l n π l n 2}{π} + \int_{0}^{\infty} \frac{l n (1 + x^{2})}{e^{2 π x} + 1} d x$

Question Number 63844 Answers: 3 Comments: 3

∫(1+4x+x^2 )^m dx

$\int {(1 + 4 x + x^{2})}^{m} d x$

Question Number 63822 Answers: 0 Comments: 1

find ∫ (x^2 +1)(√((x+1)/(x−2)))dx

$f i n d \int (x^{2} + 1) \sqrt{\frac{x + 1}{x - 2}} d x$

Question Number 63782 Answers: 1 Comments: 4

let f(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^3 )) with a>0 1) calculate f(a) 2)calculste also g(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^4 )) 3) find the values of integrals ∫_0 ^∞ (dx/((x^2 +1)^3 )) ∫_0 ^∞ (dx/((x^2 +2)^4 ))

$l e t f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{3}} w i t h a > 0$ $1) c a l c u l a t e f (a)$ $2) c a l c u l s t e a l s o g (a) = \int_{- \infty}^{+ \infty} \frac{d x}{{(a^{2} + x^{2})}^{4}}$ $3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{3}}$ $\int_{0}^{\infty} \frac{d x}{{(x^{2} + 2)}^{4}}$

Question Number 63823 Answers: 0 Comments: 0

calculate ∫_(−∞) ^(+∞) ((3x^2 −1)/(x^4 −2x^2 +3))dx

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{3 x^{2} - 1}{x^{4} - 2 x^{2} + 3} d x$

Question Number 63748 Answers: 0 Comments: 0

Question Number 63738 Answers: 0 Comments: 0

useful formula ======== ∀a∈R^+ :∀b ∈R: a sin x +b cos x =(√(a^2 +b^2 ))sin (x+arctan (b/a)) ∫(dx/(a sin x +b cos x))= =(1/(√(a^2 +b^2 )))∫(dx/(sin (x+arctan (b/a))))= [t=x+arctan (b/a) → dx=dt] (1/(√(a^2 +b^2 )))∫(dt/(sin t))=−(1/(√(a^2 +b^2 )))ln ((1/(sin t))+(1/(tan t))) = =−(1/(√(a^2 +b^2 )))ln ∣(((√(a^2 +b^2 ))−b sin x +a cos x)/(a sin x +b cos x))∣ +C

$useful formula$ $========$ $\forall a \in R^{+} : \forall b \in R : a \sin x + b \cos x = \sqrt{a^{2} + b^{2}} \sin (x + \arctan \frac{b}{a})$ $\int \frac{d x}{a \sin x + b \cos x} =$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{d x}{\sin (x + \arctan \frac{b}{a})} =$ $[t = x + \arctan \frac{b}{a} \to d x = d t]$ $\frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{d t}{\sin t} = - \frac{1}{\sqrt{a^{2} + b^{2}}} \ln (\frac{1}{\sin t} + \frac{1}{\tan t}) =$ $= - \frac{1}{\sqrt{a^{2} + b^{2}}} \ln ∣ \frac{\sqrt{a^{2} + b^{2}} - b \sin x + a \cos x}{a \sin x + b \cos x} ∣ + C$

Question Number 63722 Answers: 1 Comments: 6

1) calculate ∫ (x^2 −x+2)(√(x^2 −x+1))dx 2)find the value of ∫_0 ^1 (x^2 −x+2)(√(x^2 −x +1))dx .

$1) c a l c u l a t e \int (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x$ $2) f i n d t h e v a l u e o f \int_{0}^{1} (x^{2} - x + 2) \sqrt{x^{2} - x + 1} d x .$

Question Number 63721 Answers: 1 Comments: 2

calculate ∫ (dx/(√((x−1)(2−x))))

$c a l c u l a t e \int \frac{d x}{\sqrt{(x - 1) (2 - x)}}$

Question Number 63720 Answers: 1 Comments: 2

calculate ∫(√((x−3)(2−x)))dx

$c a l c u l a t e \int \sqrt{(x - 3) (2 - x)} d x$

Question Number 63711 Answers: 1 Comments: 1

calculate ∫_0 ^π (dx/((√3)cosx +(√2)sinx))

$c a l c u l a t e \int_{0}^{π} \frac{d x}{\sqrt{3} c o s x + \sqrt{2} s i n x}$

Pg 220 Pg 221 Pg 222 Pg 223 Pg 224 Pg 225 Pg 226 Pg 227 Pg 228 Pg 229

Contact: info@tinkutara.com