Please help..
ʃ (4sin²x cos²×/sin 2x cos 2x)dx
That's integration of (4sin^2x cos^2x over sin 2x cos 2x) dx
i've got it from the back and it has an answer from the back page of the book but i want to know how to solve it..i really need it for practicing this subject...please please help..Thanks a lot..
the answer is: -1/2 ln|secx + tanx| + c
4sin^2(x)cos^2(x)=(sin2x)^2
(sin2x/cos2x)dx=(-1/2)d(cos2x)/cos2x
Hmm. There's something wrong here. Mgraph's solution is correct, but is not what you were expecting.
ln|secx + tanx| is the integral of secx, not sinx/cosx
Is there a typo somewhere?
To solve the integral ∫ (4sin^2x cos^2x / sin 2x cos 2x) dx, you can start by simplifying the expression inside the integral.
Since sin 2x = 2sin x cos x and cos 2x = cos^2x - sin^2x, you can rewrite the integral as:
∫ (4sin^2x cos^2x / 2sin x cos x (cos^2x - sin^2x)) dx
Next, you can cancel out the common factors in the numerator and denominator:
∫ (2sin x cos x) dx
Now, you can use the trigonometric identity sin 2θ = 2sin θ cos θ to simplify further:
∫ sin 2x dx
To integrate sin 2x, you can apply the power rule of integration. Since the derivative of -cos 2x is sin 2x, the integral of sin 2x is equal to -1/2 cos 2x.
Therefore, the integral becomes:
-1/2 ∫ cos 2x dx
Now, you can substitute u = 2x, and du = 2 dx, which simplifies the integral to:
-1/4 ∫ cos u du
Integrating cos u with respect to u gives:
-1/4 sin u + C
Finally, substitute back u = 2x:
-1/4 sin 2x + C
So, the integral ∫ (4sin^2x cos^2x / sin 2x cos 2x) dx simplifies to -1/4 sin 2x + C.
Please note that there was an error in your provided answer. The correct answer should be -1/4 sin 2x + C, where C represents the constant of integration.
To solve the integral ∫ (4sin²x cos²x)/(sin 2x cos 2x) dx, we can simplify the integrand before integrating.
First, let's simplify the expression inside the integral:
4sin²x cos²x / (sin 2x cos 2x)
Using the double-angle formulas, we can rewrite sin 2x and cos 2x as follows:
sin 2x = 2sinx cosx
cos 2x = cos²x - sin²x
Now, let's substitute these expressions back into the original integral:
∫ (4sin²x cos²x)/(2sinx cosx)(cos²x - sin²x) dx
We can cancel out some terms in the numerator and denominator:
∫ 2sinx cosx dx
Next, we can use the double-angle formula for sine to rewrite sinx cosx as follows:
sinx cosx = (1/2)sin 2x
Now we can substitute this expression back into the integral:
∫ (1/2)sin 2x dx
The integral of sin 2x is:
(1/2) ∫ sin 2x dx = -(1/4) cos 2x
Applying this result to our original integral:
∫ 2sinx cosx dx = ∫ (1/2)sin 2x dx = -(1/4) cos 2x + c
Finally, we can substitute back sin 2x and cos 2x:
-(1/4) cos 2x = -(1/4) (cos²x - sin²x) = -(1/4) cos²x + (1/4)sin²x
Combining everything together, our final result is:
-(1/4) cos²x + (1/4)sin²x + c
To simplify this further, we can use the identity cos²x = 1 - sin²x:
-(1/4) (1 - sin²x) + (1/4)sin²x = -(1/4) + (1/4)sin²x + (1/4)sin²x = -(1/2) + (1/2)sin²x
Therefore, the final answer to the integral is:
-(1/2) + (1/2)sin²x + c
Note: The answer you provided, -1/2 ln|secx + tanx| + c, does not match the solution to the given integral. It is possible that there was a mistake in the given answer or in the statement of the original integral.