Trigonometric Integrals (Part II)
∫ sin2x cos3x dx
We note that the number 3 is an odd number, and so we aim at it.
∫ sin2x cos3x dx = ∫ sin2x 1 − sin2( x)d sin x
Split our one copy of cosine x. There remain two copies of cosine x, which means that we can convert it here.
cos2x = 1 − sin2x
This is cosine x square and sin x square. One copy of cosine x becomes this one.
(u = sin sin x)
Now, by substituting this expression into the integral of a polynomial, we can derive the following formula:
∫ sin2x cos3x dx = ∫ sin2x 1 − sin2( x)d sin x
= ∫(u2 − u4)du
So we set u equal to sin x. Then, du equals -sin x dx, and this is equal to one minus du squared. Multiplying out the terms and dividing by -1 gives us an integral that can be solved immediately, giving us our answer.
∫ sin2x cos3x dx = ∫ sin2x 1 − sin2( x)d sin x
= ∫(u2 − u4)du
=13u3 −15u5+ C
Finally, we substitute the value of y in the original equation, obtaining the final formula in terms of x.
13sin3x −15sin5x + C
The third example is this one.
∫ cos4x dx
Integration of the cosine function to the power of four.
∫ cos4x dx = ∫1+ cos (2x)( 2 )2dx
Now the powers of sine and cosine are both even in this case, because this power of sine is just zero.
The power of cosine is just four. Now we convert two copies of cosine into cosine 2x.
2x =1+cos (2x)2
The double-angle formula for cosine 2x is and this takes up two copies of cosine. So the four copies become two copies and two copies, then we multiply out the square, in the bottom you get four and then we divide by two. In the numerator we get this.
∫1+2 cos (2x)+cos2(2x)4dx
Now, by integrating the fourth power of x from cosine x, we have gained something. We were able to separate the terms and integrate each separately. The first term, 1/4, gives us the answer.
∫ cos4x dx = ∫1+cos (2x)( 2 )2dx
= ∫1+2 cos (2x)+cos2(2x)4dx
=14x +14sin (2x) +1412x +18 ( sin (4x) ) + C
The second term is integrated by applying the formula for integration of cosine two x. The answer is immediately this one, and finally, we have to do integration of cosine two x squared. We can apply this method once again or use the reduction formula we learn earlier for cosine to the power n. Anyway, using such a formula, we end up with this final answer for the last part here. To solve cosine 4x, we can use the double-angle formula for cosine, and then integrate each term separately.
∫ cos2(2x)dx =12∫ 1 + cos (4x))dx
=12x +14 ( sin (4x) )
We can apply a similar technique to the combination of trigonometric functions. For example, we can evaluate the tangent x to the fourth power.
Example: Integrals of Powers of tan tan x and sec sec x
Method: use 1 + tan2x = sec2x and integrate by parts evaluate ∫ tan4x dx
We first of all split out two copies of tangent x, where the result is tangent x squared.
∫ tan4x dx = ∫ tan2x tan2x dx = ∫ tan2x (sec2x − 1) dx
Tangent x square can be transformed into secant x square minus one using this identity.
∫ tan4x dx = ∫ tan2x tan2x dx = ∫ tan2x (sec2x − 1) dx
= ∫ tan2x sec2x dx − ∫ tan2x dx
Then we can write this as two integrals. The first one is kept here and the second one we use this formula again to write as secant x squared minus one; therefore, we have this one.
∫ tan4x dx = ∫ tan2x tan2x dx = ∫ tan2x (sec2x − 1) dx
= ∫ tan2x sec2x dx − ∫ tan2x dx
∫ tan2x sec2x dx − ∫(sec2x − 1) dx
∫ tan2x sec2x dx − ∫ sec2x dx + ∫ 1 dx
For the first integral, we set u equal to tangent x. When u is equal to tangent x, du equals secant x squared dx.
u = tan tan x, du = sec2x dx
Thus, the first integral becomes u square du.
∫ u2du =13u3+ C1
Thus, this is the first integral. After integrating it immediately, we get one over three times u cubed plus a certain constant. Setting u equal to this yields the first term in our series solution.
∫ tan4x dx =13tan3x − tan x + x + C
We know this integral immediately as the tangent of x, since differentiating the tangent function gives us secant x. The last two steps are simply plugging in x = 0 and x = pi/2.
∫ sec3x tan x dx
We will consider the integral of secant x times tangent x. We begin by applying integration by parts.
∫ sec3x tan x dx = ∫ sec x tan x d tan x
We set u to be this function:
(u = sec x tan x)
Then we try to find the function v, such that ev equals secant x squared over dx.
(dv = sec2x dx)
We find that v is equal to the tangent x.Therefore, by means of integration by parts,we can write this as u times v minus v times du.
∫ sec3x tan x dx = ∫ sec x tan x d tan x
= sec x tan2x − ∫ tan x (sec3x + tan2x sec x) dx
When we differentiate secant x tangent x, we arrive at this. Making use again of the formula for tangent x square, which is equal to secant x square minus one, we can simplify the thing inside the bracket by taking two secant x cubes and subtracting one secant x cube.
∫ sec3x tan x dx = ∫ sec x tan x d tan x
= sec x tan2x − ∫ tan x (sec3x + tan2x sec x) dx
= sec x tan2x − ∫ tan x (2sec3x − sec x) dx
We have three copies of the original integral, which means it is equal to the first term here plus the integral of secant x times tangent x.
3∫ sec3x tan x dx = sec x tan2x + ∫ tan x sec x dx
And this last expression is equal to the derivative of secant x, thus the integral of this last expression is equal to x times 1/3 plus a constant. And by canceling like terms, we get that the final answer is 1/3 times secant x cubed plus a constant.
3∫ sec3x tan x dx = sec x tan2x + ∫ tan x sec x dx
= sec x tan2x + sec x + C
= sec x (tan2x + 1) + C = sec3x + C
∫ sec3x tan x dx =13sec3x + C
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