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11. x tan (x +1) sec (x+1) dx

11. x tan (x +1) sec (x+1) dx

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.6: The Matrix Of A Linear Transformation
Problem 30EQ
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Could you slove question number 11 by the mothod on the example please

In exercises 1-44, evaluate the integrals. 1. cos x sin* x dxr 2. cos' x sin x dx 7/4 T/3 3. cos 2x sin 2x dx 4. | cos 3x sin 3x dx 7/4 7/2 cos x sin x dx | cos' x sin x dx 5. 6. -1/2 | cos (x+ 1) dx cos (x + 1) dx | sin (x – 3) dx 7. 8. 9. tan x sec' x dx 10. cot x csc* x dx 11. x tan (x2 + 1) sec (x2 +1) dx 12. tan (2x + 1) sec' (2x + 1) dx 13. | cot x csct x dx 14. cot x csc2 x dx •7/4 | tan“ x sect x dx Ltan 15. 16. tan“ x sec? x dx -7/4 17. cos? x sin? x dx 18. (cos? x+ sin x) dx • /2 19. sin' x dx cot? x csc x dx 20. 1/4 cos x -1/3
Transcribed Image Text:In exercises 1-44, evaluate the integrals. 1. cos x sin* x dxr 2. cos' x sin x dx 7/4 T/3 3. cos 2x sin 2x dx 4. | cos 3x sin 3x dx 7/4 7/2 cos x sin x dx | cos' x sin x dx 5. 6. -1/2 | cos (x+ 1) dx cos (x + 1) dx | sin (x – 3) dx 7. 8. 9. tan x sec' x dx 10. cot x csc* x dx 11. x tan (x2 + 1) sec (x2 +1) dx 12. tan (2x + 1) sec' (2x + 1) dx 13. | cot x csct x dx 14. cot x csc2 x dx •7/4 | tan“ x sect x dx Ltan 15. 16. tan“ x sec? x dx -7/4 17. cos? x sin? x dx 18. (cos? x+ sin x) dx • /2 19. sin' x dx cot? x csc x dx 20. 1/4 cos x -1/3
We illustrate this in example 3.6. Integration Techniques 6-16 EXAMPLE 3.6 An Integrand with an Odd Power of Tangent Evaluate f tan'x sec'x dx. Solution Looking for terms that are derivatives of other terms, we rewrite the integral as tan x sec' x dx = tan x sec² x (sec x tan x) dx (sec? x- 1) sec?x (sec x tan x) dx, where we have used the Pythagorean identity tan? x = sec2 x - 1. You should see the substitution now. We let u = sec x, so that du = sec x tan x dx and hence, tan x sec'x dx = | (sec? x 1) sec² x (sec x tan x) dx (u? - 1)u? du = (u - 1)u°du = [u* - u') du 1 u'+c = 1 sec x - 1 +c=sec'x- sec'x+ c. Since u = sec x Case 2: n Is an Even Positive Integer First, isolate one factor of sec? x. (You'll need this for du.) Then, replace any remaining factors of sec2 x with 1+ tan2 x and make the substitution u = tan x. We illustrate this in example 3.7. EXAMPLE 3.7 An Integrand with an Even Power of Secant Evaluate / tan?x sec x dx. Solution Since tan x sec? x, we rewrite the integral as tan? x sec* x dx = tan? x sec? x sec2 x dx = tan? x(1 + tan? x) sec2 x dx. Now, we let u = tan x, so that du sec? x dx and | tan? x sec* x dx = tan? x(1 + tan? x) sec? x dx du u²(1+u²) + u²)du = | (u² + u*) du 1 + -u' + c 1 tan' x + 1 tan x+ c. Since u = tan x. SECTION 6.3 Trigonometric Techniques of Integration 463 .. Case 3: m Is an Even Positive Integer and n Is an Odd Positive Integer Ronlooo on fton2 v ..ith tiou fouula (giuon 15
Transcribed Image Text:We illustrate this in example 3.6. Integration Techniques 6-16 EXAMPLE 3.6 An Integrand with an Odd Power of Tangent Evaluate f tan'x sec'x dx. Solution Looking for terms that are derivatives of other terms, we rewrite the integral as tan x sec' x dx = tan x sec² x (sec x tan x) dx (sec? x- 1) sec?x (sec x tan x) dx, where we have used the Pythagorean identity tan? x = sec2 x - 1. You should see the substitution now. We let u = sec x, so that du = sec x tan x dx and hence, tan x sec'x dx = | (sec? x 1) sec² x (sec x tan x) dx (u? - 1)u? du = (u - 1)u°du = [u* - u') du 1 u'+c = 1 sec x - 1 +c=sec'x- sec'x+ c. Since u = sec x Case 2: n Is an Even Positive Integer First, isolate one factor of sec? x. (You'll need this for du.) Then, replace any remaining factors of sec2 x with 1+ tan2 x and make the substitution u = tan x. We illustrate this in example 3.7. EXAMPLE 3.7 An Integrand with an Even Power of Secant Evaluate / tan?x sec x dx. Solution Since tan x sec? x, we rewrite the integral as tan? x sec* x dx = tan? x sec? x sec2 x dx = tan? x(1 + tan? x) sec2 x dx. Now, we let u = tan x, so that du sec? x dx and | tan? x sec* x dx = tan? x(1 + tan? x) sec? x dx du u²(1+u²) + u²)du = | (u² + u*) du 1 + -u' + c 1 tan' x + 1 tan x+ c. Since u = tan x. SECTION 6.3 Trigonometric Techniques of Integration 463 .. Case 3: m Is an Even Positive Integer and n Is an Odd Positive Integer Ronlooo on fton2 v ..ith tiou fouula (giuon 15
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