Find anti-derivative of

Question

An illustration that pairs with a mathematics problem. The image should visualize the integral calculus. It should include a graph showing sine and cosine waves, with a focus on the curves and peaks that represent the integral of (the square of the sum of the sine of x with the triple of the cosecant of x) plus the square of the cosine of x. The image must be clear, simple, and educational, fully devoid of text.

Find anti-derivative of

(sinx + 3cosecx)^2 + cos^2x

Answer 1

(sinx + 3cscx)^2 + cos^2x

= sin^2x + 6 + 9csc^2x + cos^2x
= 7 + 9csc^2x

∫ 7+9csc^2x dx = 7x - 9cotx + C

Answer 2

Why did the scarecrow win an award? Because he was outstanding in his field! Now, let's find the anti-derivative of (sinx + 3cosecx)^2 + cos^2x:

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!

Answer 3

To find the antiderivative of the given expression, let's simplify it first.

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.

Answer 4

To find the antiderivative of the given expression, let's break it down into two separate terms:

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.