T * sin ^ 3 (T) DT

T * sin ^ 3 (T) DT

sin^3(t) = [1-cos(2t)] sint / 2 = (3/4)sint - (1/4) sin(3t)
I = (3/4) ∫ t d(-cost) + (1/12) ∫ t d(cos3t)
= (-3/4) [ t cost - ∫ cost dt ] + (1/12) [ t cos(3t) - ∫ cos(3t) dt ]
= (-3/4) [ t cost - sint] + (1/12) [ t cos(3t) - (1/3) sin(3t) ] + C

How to calculate the integral sign [1 / sin ^ 2 (T)] DT

Original = - ∫ (- CSC & # 178; t) DT
=-cott+C

The integral process of sin (x squared),

This function is not integrable, that is, the result cannot be expressed as elementary function. It can be calculated as follows: SiNx = ∑ (n: 1 →∞) (- 1) ^ (n-1) * x ^ (2n-1) / (2n-1)! Sin (x ^ 2) = ∑ (n: 1 →∞) (- 1) ^ (n-1) * (x ^ 2) ^ (2n-1) / (2n-1)! = ∑ (n: 1 →∞) (- 1) ^ (n-1) * x ^ (4n-2) / (2n-1)

What is the integral of sin squared x

The voltage formula is u (x) = 310sin square (x) \ \ x0d, and then calculate the definite integral of U (x) from 0 to t, and then divide it by t \ \ x0d to get the effective value of voltage

What is the integral of (cos2x) / sin square x?

Original formula = integral (COS squared x-sin squared x) / sin squared x = integral (1-2 sin squared x) / sin squared x = integral [(1 / sin squared x) - 2] = - cot x-2x + C

How to integrate sin 5x ^ 2? How to integrate the square of sin 5x
Sin5 (x ^ 2) is not sin ^ 2 (5x)

Double angle formula
cos10x=1-2(sin5x)^2
∫(sin5x)^2dx
=∫(1-cos10x)/2dx
=∫1/2 dx-∫cos10x dx
=x/2-(sin10x)/10+C

Let f (x) = LIM (Sint / SiNx) ^ X / Sint SiNx determine the breakpoints and point out the types
T → X!

When sin / SiN x approaches 1, sin / SiN x = 1 + [(sin / SiN x) - 1] becomes ln (1 + y) type. When sin / SiN x approaches 1, y approaches 0, and ln (1 + y) approaches y, that is, (sin / SiN x) - 1. The first kind of discontinuities include de discontinuities and jumps

(∫ (0 to x) t * e ^ t * Sint DT) / x ^ 6 * e ^ x, find the limit, X tends to 0
Is that right?
=2x*x^2*e^(x^2)*sinx^2/6x^5*e^x+x^6*e^x
=2x^5*e^(x^2)/e^x(6x^5+x^6)
=2x^5{(e^(x^2)-1) +1}/{(e^x-1)+1}(6x^5+x^6)
=2x^5(x^2+1)/(x+1)(6x^5+x^6)
=2x^7+2x^5/7x^6+x^7+6x^5
Since x tends to 0, x ^ 7 and x ^ 6 are infinitesimals of higher order of x ^ 5, so
=2x^5/6x^5=1/3
(∫ (0 to x ^ 2) t * e ^ t * Sint DT) / x ^ 6 * e ^ x, find the limit, X tends to 0, just wrong

The integral interval is (0, x ^ 2)
The above steps are all right, so we can simplify them
=LIM (x - > 0) 2x ^ 5 * e ^ (x ^ 2) / e ^ x (6x ^ 5 + x ^ 6) x - > 0, e ^ x ^ 2 / e ^ x = e ^ (x ^ 2-x) - > 1
=lim(x->0)2x^5/(6x^5+x^6)
=1/3

lim x→0[(∫(x,0)(x-cost)dt)/x^3]=?

Integral (∫ (x, 0) (x-cost) DT = (XT Sint) | (x, 0) = x ^ 2-sinx
Then use the law of Robida limx - > 0 (x ^ 2-sinx) / x ^ 3 = limx - > 0 (2x cosx) / 3x ^ 2 = ∞

lim x→∞(∫（x,0)｜cost｜dt)/x=?

|Cost | is a positive function with a period of Pie / 2
And it is easy to know that the integrals on (0, Pie / 2) and (Pie / 2, pie) and (n * Pie / 2, (n + 1) * Pie / 2) are the same
And (0, Pie / 2) integral is equal to 1
Let x = n * Pie / 2 + K (where 0=