(sinx + 3cscx)^2 + cos^2x
= sin^2x + 6 + 9csc^2x + cos^2x
= 7 + 9csc^2x
∫ 7+9csc^2x dx = 7x - 9cotx + C
(sinx + 3cosecx)^2 + cos^2x
= sin^2x + 6 + 9csc^2x + cos^2x
= 7 + 9csc^2x
∫ 7+9csc^2x dx = 7x - 9cotx + C
First, let's expand and simplify the expression:
(sin^2x + 2sinx*3cosecx + (3cosecx)^2) + cos^2x
(sin^2x + 6sinx*cosecx + 9cosec^2x) + cos^2x
Now, let's use the trigonometric identity: cosec^2x = 1 + cot^2x
(sin^2x + 6sinx*cosecx + 9(1 + cot^2x)) + cos^2x
sin^2x + 6sinx*cosecx + 9 + 9cot^2x + cos^2x
Simplifying further:
(sin^2x + cos^2x) + 6sinx*cosecx + 9 + 9cot^2x
1 + 6sinx*cosecx + 9 + 9cot^2x
Now, let's find the anti-derivative of this expression:
∫(1 + 6sinx*cosecx + 9 + 9cot^2x) dx
Integrating term by term, we get:
x + 6∫sinx*cosecx dx + 9x + 9∫cot^2x dx
Now, let's simplify further:
x + 6∫sinx*(1/sinx) dx + 9x + 9∫cot^2x dx
x + 6∫dx + 9x + 9∫cot^2x dx
x + 6x + 9x + 9∫cot^2x dx
16x + 9∫cot^2x dx
Now, the anti-derivative of cot^2x is -cotx:
16x - 9cotx + C
So, the anti-derivative of (sinx + 3cosecx)^2 + cos^2x is 16x - 9cotx + C, where C is the constant of integration. Now, that's a math problem solved with a touch of clown humor!
The given expression is:
(f(x))^2 + g(x)
Let's expand (f(x))^2 using the binomial theorem:
(f(x))^2 = (sinx + 3cosecx)^2
= sin^2x + 2sinx(3cosecx) + (3cosecx)^2
= sin^2x + 6sinxcosxsecx + 9cosec^2x
Now, the given expression becomes:
(sin^2x + 6sinxcosxsecx + 9cosec^2x) + cos^2x
Combining like terms:
= sin^2x + cos^2x + 6sinxcosxsecx + 9cosec^2x
Recall the Pythagorean identity: sin^2x + cos^2x = 1. So we can simplify the expression further:
= 1 + 6sinxcosxsecx + 9cosec^2x
Now, let's find the antiderivative of each term one by one.
The antiderivative of 1 with respect to x is x.
The antiderivative of sinxcosxsecx can be found using the substitution method.
Let's substitute u = sinx.
Then, du = cosxdx.
The integral becomes:
∫ du/secx
Using the identity: secx = 1/cosx, the integral becomes:
∫ du/(1/cosx)
= ∫cosx du
= sinx + C1, where C1 is the constant of integration.
The antiderivative of 9cosec^2x can be found using the identity:
d(cotx) = -cosec^2x dx
Hence, -9 ∫ d(cotx) = -9cotx + C2, where C2 is the constant of integration.
Therefore, the antiderivative of the given expression is:
x + sinx + 6sinxcosxsecx + 9cosec^2x + C, where C = C1 + C2, is the constant of integration.
1. The antiderivative of (sin(x) + 3csc(x))^2:
First, expand the square of the expression:
(sin(x) + 3csc(x))^2 = sin^2(x) + 6sin(x)csc(x) + 9csc^2(x)
To integrate sin^2(x), we can use the identity:
sin^2(x) = (1 - cos(2x))/2
Therefore, the integral of sin^2(x) is:
∫(1 - cos(2x))/2 dx = x/2 - (sin(2x))/4 + C1, where C1 is the constant of integration.
To integrate 6sin(x)csc(x), we can use the substitution u = sin(x):
Then, du = cos(x) dx.
Rewrite 6sin(x)csc(x) as 6(u)/(u^2 - 1) and make the substitution:
6∫(u)/(u^2 - 1) du
Using partial fraction decomposition, we can write:
6∫(u)/(u^2 - 1) du = 3∫(1/(u - 1) - 1/(u + 1)) du
= 3(ln|u - 1| - ln|u + 1|) + C2, where C2 is the constant of integration.
Finally, to integrate 9csc^2(x), we can use the substitution u = cos(x):
Then, du = -sin(x) dx.
Rewrite 9csc^2(x) as -9(u^2 + 1)/(u^2 - 1) and make the substitution:
-9∫(u^2 + 1)/(u^2 - 1) du
Using partial fraction decomposition, we can write:
-9∫(u^2 + 1)/(u^2 - 1) du = -9∫(1/(u - 1) + 1/(u + 1)) du
= -9(ln|u - 1| + ln|u + 1|) + C3, where C3 is the constant of integration.
Putting it all together, the antiderivative of (sin(x) + 3csc(x))^2 is:
x/2 - (sin(2x))/4 + 3(ln|sin(x) - 1| - ln|sin(x) + 1|) - 9(ln|cos(x) - 1| + ln|cos(x) + 1|) + C, where C = C1 + C2 + C3 is the constant of integration.
2. The antiderivative of cos^2(x):
Recall the power-reducing formula: cos^2(x) = (1 + cos(2x))/2
Therefore, the integral of cos^2(x) is:
∫(1 + cos(2x))/2 dx = x/2 + (sin(2x))/4 + C4, where C4 is the constant of integration.
The antiderivative of the given expression (sin(x) + 3csc(x))^2 + cos^2(x) is:
x/2 - (sin(2x))/4 + 3(ln|sin(x) - 1| - ln|sin(x) + 1|) - 9(ln|cos(x) - 1| + ln|cos(x) + 1|) + x/2 + (sin(2x))/4 + C, where C is the constant of integration.