Sorry, i have a load of questions on integration... thanks for any help provided!
Evaluate the integrals:
limit 0 to pi/4 ∫ [sec^2x]/[5+tanx] dx
limit 0 to pi/6 ∫ [3cos3x]/[3+sin3x] dx
limit 0 to 3 ∫ [2x-1]/[x^2-x+1] dx
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By squaring sin^2 + cos^2, show that sin^4x + cos^4x = 1/4 (3+cos4x)
Hence, evaluate ∫(sin^4x+cos^4x) dx
(limit from 0 to pi/8)
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How do i express (2x+1)/(x+3) in the form of p + q/(x+3)???
Sure! I'll help you with your questions on integration.
1. Evaluate the integral:
To evaluate the integral limit 0 to π/4 ∫ [sec^2x]/[5+tanx] dx, you can start by simplifying the expression:
sec^2x = 1 + tan^2x
The integral becomes:
∫ [1 + tan^2x]/[5 + tanx] dx
Now, you can make a substitution: let u = tanx, du = sec^2x dx
This changes the integral to:
∫ [1 + u^2]/[5 + u] du
Now, you can use partial fraction decomposition or simplify further to evaluate this integral.
2. Evaluate the integral:
To evaluate the integral limit 0 to π/6 ∫ [3cos3x]/[3+sin3x] dx, you can use the substitution method. Let u = 3x, du = 3dx
The integral becomes:
1/3 ∫ [3cosu]/[3+sinu] du
Now, you can use partial fraction decomposition or simplify further to evaluate this integral.
3. Evaluate the integral:
To evaluate the integral limit 0 to 3 ∫ [2x-1]/[x^2-x+1] dx, you can use the partial fraction decomposition method if the denominator can be factored.
To factor x^2 - x + 1, you can complete the square or use the quadratic formula. After factoring, you can find the partial fraction representation and evaluate the integral.
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By squaring sin^2x + cos^2x, you get:
(sin^2x + cos^2x)^2 = sin^4x + 2sin^2xcos^2x + cos^4x
Now, using the trigonometric identity sin^2x + cos^2x = 1, you can substitute it into the equation above:
(1)^2 = sin^4x + 2sin^2xcos^2x + cos^4x
This simplifies to:
1 = sin^4x + cos^4x + 2sin^2xcos^2x
Rearranging the terms, you get:
sin^4x + cos^4x = 1 - 2sin^2xcos^2x
Now, using the identity sin2x = 2sinxcosx, you can substitute it into the equation above:
sin^4x + cos^4x = 1 - (sin2x)^2
The right side can be further simplified using the identity sin^22x + cos^22x = 1:
sin^4x + cos^4x = 1 - (1 - cos^22x)/2
simplifying further gives:
sin^4x + cos^4x = 1/4 (3 + cos4x)
Therefore, ∫(sin^4x + cos^4x) dx from 0 to π/8 = ∫(1/4 (3 + cos4x)) dx from 0 to π/8 = 1/4 ∫ (3 + cos4x) dx from 0 to π/8
Evaluate this integral using the given limits to find the result.
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To express (2x+1)/(x+3) in the form of p + q/(x+3), you can use long division or synthetic division.
Perform (2x+1) divided by (x+3) to get the quotient p and the reminder q. The quotient will involve the "p" term, and the remainder will be the "q" term.
For example, using long division:
2x + 1
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x + 3 | 2x + 1
Dividing, you get the quotient as 2 and the remainder as -5. So, (2x+1)/(x+3) = 2 + (-5)/(x+3), which is in the desired form p + q/(x+3).
I hope this helps! If you have any further questions, feel free to ask.