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The integral calculus is concerned with the inverse problem of the given derivative of a function to find the function. In symbol, we require to find f(x) where, d/dx f(x) = g(x) and g(x) is given.Then f(x) = ? g(x) dx. Thus, Define the integration as follows: The integral of the function g(x) with respect to x is the function whose derivative with respect to x is g(x) and is written as = ? g(x) dx.
Basic Integral Formula in Trigonometric Integrals:
1. ? x n dx = (xn+1) / (n+1)
2. ? cos x .dx= sin x. + c
3. ? sin x.dx = - cos x + c
4. ? sec2x = tan x + c
5. ? cosec x.cot x = -cosec x + c
6. ? sec x.tan x = sec x + c
7. ? cosec2 x = -cot x + c
8. ?(1/x) dx = log x + c
9. ? e x dx = e x + c
10. ? ( 1/(1+ x 2) ) dx = tan -1 x + c
Trigonometric Integrals Examples
1. Integrate the given function with respect to x: ? sin2(x /2).dx
Solution:
? sin2(x /2).dx = ? (1/2) (1- cos x) dx ; we know, sin2 (x /2) = (1- cos x) / 2
= ?(1/2) 1 dx - ?(1/2)cos x dx
= (1/2) ?dx - (1/2) ?cos x dx ; ?cos x dx = sin x + c
= (1/2) x - (1/2) sin x + c
Answer:
? sin2(x /2).dx = (1/2) x - (1/2) sin x + c
2. Integrate: ?cos42x dx
Solution:
?cos42x dx = ? (cos22x)2dx
= ? ((1+ cos 4x) / 2)2dx
= (1/4) ? ( 1+2 cos 4x + cos24x) dx
= (1/4) ? ( 1+2 cos 4x + (( 1+ cos 8x ) /2)) dx
= (1/8)? (2 + 4 cos 4x + 1+ cos 8x) dx
= (1/8) ? (3 + 4 cos 4x + cos 8x) dx
= (1/8) [3x + ((4 sin 4x) / 4) + ((sin 8x ) / 8)] + c
= (1/64) [24x + 8 sin 4x + sin 8x] +c
Answer:
?cos42x dx = (1/64) [24x + 8 sin 4x + sin 8x] +c
Trigonometric integral practice problem:
1. Integrate: ? (sin6x dx / cos8x) dx
Answer: (1/7) (tan x)7 + c
2. Integrate: ? (1/x) sec2(log x) dx
Answer: tan ( log x) + c
Students can also get help with Calculus homework problems from the online tutors.