Problem to solve
Find the indefinite integral \[\int\tan^4[x]\sec^5[x] dx\]
Solution
STEP 1
Take the integral: integral tan^4(x) sec^5(x) dx
STEP 2
For the integrand tan^4(x) sec^5(x), use the trigonometric identity tan^2(x) = sec^2(x) - 1: = integral sec^5(x) (sec^2(x) - 1)^2 dx
STEP 3
Expanding the integrand sec^5(x) (sec^2(x) - 1)^2 gives sec^9(x) - 2 sec^7(x) + sec^5(x): = integral(sec^9(x) - 2 sec^7(x) + sec^5(x)) dx
STEP 4
Integrate the sum term by term and factor out constants: = integral sec^9(x) dx - 2 integral sec^7(x) dx + integral sec^5(x) dx
STEP 5
Use the reduction formula, integral sec^m(x) dx = (sin(x) sec^(m - 1)(x))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(x) dx, where m = 9: = 1/8 tan(x) sec^7(x) - 9/8 integral sec^7(x) dx + integral sec^5(x) dx
STEP 6
Use the reduction formula, integral sec^m(x) dx = (sin(x) sec^(m - 1)(x))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(x) dx, where m = 7: = -3/16 tan(x) sec^5(x) + 1/8 tan(x) sec^7(x) + 1/16 integral sec^5(x) dx
STEP 7
Use the reduction formula, integral sec^m(x) dx = (sin(x) sec^(m - 1)(x))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(x) dx, where m = 5: = 1/64 tan(x) sec^3(x) - 3/16 tan(x) sec^5(x) + 1/8 tan(x) sec^7(x) + 3/64 integral sec^3(x) dx
STEP 8
Use the reduction formula, integral sec^m(x) dx = (sin(x) sec^(m - 1)(x))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(x) dx, where m = 3: = 3/128 tan(x) sec(x) + 1/64 tan(x) sec^3(x) - 3/16 tan(x) sec^5(x) + 1/8 tan(x) sec^7(x) + 3/128 integral sec(x) dx
STEP 9
The integral of sec(x) is log(tan(x) + sec(x)): = 1/8 tan(x) sec^7(x) - 3/16 tan(x) sec^5(x) + 1/64 tan(x) sec^3(x) + 3/128 tan(x) sec(x) + 3/128 log(tan(x) + sec(x)) + constant
STEP 10
Factor the answer a different way: = 1/128 (16 tan(x) sec^7(x) - 24 tan(x) sec^5(x) + 2 tan(x) sec^3(x) + 3 tan(x) sec(x) + 3 log(tan(x) + sec(x))) + constant
STEP 11
Which is equivalent for restricted x values to: Answer: | | = 3/128 tanh^(-1)(sin(x)) - 3/40 tan(x) sec^7(x) + 1/5 tan^3(x) sec^5(x) + 1/80 tan(x) sec^5(x) + 1/64 tan(x) sec^3(x) + 3/128 tan(x) sec(x) + constant
More problems
Number
001Find the indefinite integral \(\int\frac{x^2+2}{x^3+5x^2+2x-8} dx\)
002Find the indefinite integral \(\int\frac{4x^4-2}{x^3+4x^2+x-6} dx\)
003Find the indefinite integral \(\int\frac{4x^4-22x^3+11x^2-99x-34}{x^3-6x^2+4x-24} dx\)
004Find the indefinite integral \(\int\frac{3x^4+4x^2+5x}{(x-1)(x^2+1)^2} dx\)
005Find the indefinite integral \(\int\frac{x^6+6x^4+8x^2-4}{(x^2+4)(x^2+3)} dx\)
006Find the indefinite integral \(\int(-4x^4+6x^2+8x-1)\cos(x) dx\)
007Find the indefinite integral \(\int(-7x^5+2x^4-2)\cdot3^xdx\)
008Find the indefinite integral \(\int\arcsin^4(x) dx\)
009Find the indefinite integral \(\int\sin(\ln(x)) dx\)
010Find the indefinite integral \(\int(12x^{15}-3x^7-4x^3)\cdot e^{2x^4} dx\)
011Find the indefinite integral \(\int\frac{\sqrt{(3x^2+4)^3}}{x^4} dx\)
012Find the indefinite integral \(\int\frac{x^2}{\sqrt{2x^2-5}} dx\)
013Find the indefinite integral \(\int\frac{x^6}{\sqrt{(x^2+4)^3}} dx\)
014Find the indefinite integral \(\int x^2\sqrt{4-3x^2} dx\)
015Find the indefinite integral \(\int x^4\cdot\sqrt{x^2+4x+5} dx\)
016Find the indefinite integral \(\int\frac{\sqrt{x+2}}{x+1} dx\)
017Find the indefinite integral \(\int\frac{\sqrt{x+2}-4}{5-\sqrt{x+2}} dx\)
018Find the indefinite integral \(\int\frac{3\sqrt[3]{2x-1}-4}{\sqrt[6]{2x-1}-1} dx\)
019Find the indefinite integral \(\int\sqrt{\frac{x+3}{1-x}} dx\)
020Find the indefinite integral \(\int\frac{\sqrt[3]{x+1}-\sqrt{x+1}}{x} dx\)
021Find the indefinite integral \(\int\frac{\sin(x)-3\cos(x)}{\cos(x)-2} dx\)
022Find the indefinite integral \(\int\frac{3\sec(x)-\cot(x)}{1-\csc(x)} dx\)
023Find the indefinite integral \(\int\frac{\tan^2(x)+2}{\tan(x)+2} dx\)
024Find the indefinite integral \(\int\frac{3\sin^4(x)-2\cos^2(x)}{1+\sin^2(x)} dx\)
025Find the indefinite integral \(\int\tan^4(x)\sec^5(x) dx\)
Back
Edited by tododemate® - 2021