Trigonometric Substitution Alvin Lin Calculus II: August 06 - December 06 Trigonometric Substitution sin 4 (x) cos (x) dx When you have a product of sin and cos of different powers, you have three different possibilities: They are both even powers. They are both odd powers. One exponent is odd and the other is even. We can rewrite this problem as: sin 4 (x) cos (x) cos(x) dx We want even powers of sin 4 (x)( sin (x)) (x) cos(x) dx Now we can use substitution: cos(x) = Let : sin(x) = t dt dx cos(x) dx = dt t 4 ( t) dt
t 4 t 6 dt = t 4 dt t 6 dt t 5 5 t7 7 sin 5 (x) sin7 (x) 5 7 And in the wise words of Professor Khan: These terms are like Hillary and Trump supporters and we cannot combine them. Another case is where we have a difficult term inside a radical in the denominator. dx x 4 x Note that the term in the radical has a form similar to the trigonometric identities above. Let : x = sin() dx = cos() d cos() d ( sin()) 4 4 sin () By substituting for sin(), we can turn the radical into the form of a trigonometric identity. cos() 4 sin () 4 sin () d In this case, we are using the identity sin () + cos () = which we can rewrite as cos () = sin (). cos 4 sin () cos () d cos 4 sin ()cos() d 4 sin () d csc () d 4 4 cot()
To substitute back, we must imagine a triangle with angle. Given our first subsitution x = sin(), we can rewrite it as sin() = x = opp. If our triangle has opposite hyp side x and hypotenuse, then the adjacent side must be 4 x. x 4 x Therefore: cot() = adj = 4 x opp x 4 cot() = 4 x 4 x = 4 x 4x Practice Problem 4 x x dx Let : x = sin() dx = cos() d ( sin()) cos() d ( sin()) 7 sin () cos() sin () d 7 sin () cos() sin () d sin () cos() cos () sin () cos() cos() d d
sin () d Using the double angle formulas: cos() d Using the double angle formulas again: = cos() d sin() sin() cos() sin() cos() Recall that we substituted x = sin(), which we can rewrite as sin() = x = opp hyp. If we imagine a triangle in which the opposite side is x and the hypotenuse is, then the adjacent side must be x. x x Therefore: cos() = adj hyp = x and = sin ( x ) sin() cos() = sin ( x ) x = sin ( x ) x x x 4
Practice Problem 6 0 x 6 x dx Let : x = 6 sin() dx = 6 cos() d For now, we will solve the problem as an indefinite integral. 6 sin() 6 cos() d 6 (6 sin()) 6 sin() cos() 6 6 sin () d 6 sin() cos() 6 sin () d sin() cos() 6 cos () sin() cos() 6 cos() 6 sin() d 6 cos() Recall that we substituted x = 6 sin(), which we can rewrite as sin() = x 6 = opp hyp. If we imagine a triangle in which the opposite side is x and the hypotenuse is 6, then the adjacent side must be 6 x. d d 6 x 6 x 5
Therefore: cos() = adj = 6 x hyp 6 6 cos() = 6 6 x 6 = 6 x Now we can use the original limits of the intergral to solve this. 6 x 0 Practice Problem 6 ( 6 0 ) 7 + 6 = 6 dt t t 6 Note that the radical is of the form t 6. We cannot subsitute sin() into this since it will not satisfy the trigonometric identity. Let : x = 4 sec() dx = 4 sec() tan() d 4 sec tan() 6 sec () (4 sec()) 6 d() tan() 4 sec() 6 sec () d() tan() 6 sec() tan () d() tan() 6 sec() tan() d() cos() d 6 6
6 sin() Recall that we substituted x = 4 sec(), which we can rewrite as sec() = x 4 = hyp adj. If we imagine a triangle in which the hypotenuse is x and the adjacent side is 4, then the opposite side must be 6 x. x 6 x 4 Therefore: sin() = opp = 6 x hyp 4 6 sin() = 6 x 6 4 6 x 64 Practice Problem 5 x a x dx Let : x = a sin() d dx = a cos() d a sin () a a sin ()a cos() d a 4 a 4 sin () cos() sin () d sin () cos() cos () d a 4 a 4 sin () cos () d (sin() cos()) d 7
Using the double angle formulas: a 4 ( sin() ) d a 4 sin () d 4 Using the double angle formulas again: a 4 cos(4) d 4 a 4 cos(4) d a 4 4 sin(4) a 4 4 sin() cos() a 4 ( sin() cos())( sin ()) a 4 sin() cos() sin () sin() cos() a 4 sin() cos() sin () cos() Recall that we substituted x = a sin(), which we can rewrite as sin() = x a = opp hyp. If we imagine a triangle in which the opposite side is x and the hypotenuse is a, then the adjacent side must be a x. a x a x
Given this information: = sin ( x a ) sin() = x a a x cos() = a We can substitute this back into our solution: a 4 sin() cos() sin () cos() a 4 sin ( x a ) x a x x a x a a 4 Practice Problem x 5x dx The terms inside the radical are not of the same form as the problems before. We can rewrite this problem to figure out the substitution. x (5x) dx Let : 5x = sin(t) 5 dx = cos(t) dt dx = cos(t) dt 5 ( 5 sin(t)) cos(t) dt sin (t) 5 7 5 cos(t) sin (t) dt cos(t) sin (t) dt 5 ( cos(t)) dt 5
50 50 t 50 t sin(t) sin(t) cos(t) t sin(t) cos(t) Recall that we substituted 5x = sin(t), which we can rewrite as sin(t) = 5x = opp. hyp If we imagine a triangle in which the opposite side is 5x and the hypotenuse is, then the adjacent side is 5x. 5x t 5x Therefore: cos(t) = adj = 5x and t = sin ( 5x) hyp t sin(t) cos(t) 50 50 50 Practice Problem 7 sin ( 5x ) 5x 5x sin ( 5x ) 5x 5x x + x dx This problem requires a different approach. We need to turn this into the form of a trigonometric substitution problem. x x + x dx = + x + ( ) dx x + x + dx 0
(x + ) dx Let : x + = sec() dx = sec() tan() d sec () sec() tan() d Now we use integration by parts: tan () sec() tan() d tan () sec() d Let : f(x) = tan() g (x) = tan() sec() d f (x) = sec () d g(x) = sec() tan () sec() d = tan() sec() sec () sec() d tan () sec() d = tan() sec() (tan () + ) sec() d tan () sec() d = tan() sec() tan () sec() d() + sec() d tan () sec() d = tan() sec() sec() d tan () sec() d = tan() sec() ln sec(x) + tan(x) tan () sec() d = tan() sec() ln sec(x) + tan(x) Recall that we substituted x+ = sec(), which we can rewrite as sec() = x+ If we imagine a triangle in which the hypotenuse is x + and the adjacent side is, then the opposite side must be x. = hyp. adj x + x
Therefore: tan() = opp = x adj tan() sec() ln sec(x) + tan(x) = x(x + ) = ln x + x + x(x + ) ln (x + ) + x Practice Problem 45 x + x dx For this problem, we can solve it with regular substitution: x x + x dx Let : + x = t x dx = dt (t )t dt t t dt 5 t = 5 t But we can also use trigonometric substitution: x + x dx Let : x = tan() dx = sec () d tan () + tan () sec () d tan () sec () sec () d
tan () sec () d tan() tan () sec () d tan()(sec ) sec () d tan() sec 5 () d tan() sec () d tan() sec() sec 4 () d tan() sec() sec () d Let : u = sec() du = sec() tan() d u 4 du u du u 5 5 u sec 5 () sec () 5 Recall that we substituted x = tan(), which we can rewrite as tan() = x = opp. If adj we imagine a triangle in which the opposite side is x and the adjacent side is, then the hypotenuse must be + x. + x x Therefore: sec() = hyp adj = +x. sec 5 () 5 sec () = 5 ( + x ) 5 ( + x ) = ( + x ) 5 5 ( + x )
Follow-up Questions 4 6 x x x dx x (x + x ) dx x (x + x + 4 4) dx x (x + x + ) + 4 dx x 4 (x + ) dx Let : x + = cos() dx = sin() d ( cos() ) 4 ( cos()) ( sin()) d ( cos() ) 4 cos ()( sin()) d 4 ( cos() ) sin ()( sin()) d 4 (4 cos () 4 cos() + )( sin ()) d 4 4 cos () sin () 4 cos() sin () + sin () d cos () sin () d 6 ( cos () sin ()) d 6 4 sin () d 6 cos() sin () d 4 cos() sin () d cos() sin () d sin () d 6 cos () cos(4) d 6 cos () 4 sin () d sin () d cos() d ( sin()) ( sin())
sin(4) 6 cos () ( sin()) sin(4) 6 cos () + sin() sin() sin(4) 6 cos () sin() cos() sin() cos() 6 cos () sin() cos() ( sin() cos())( cos () ) 6 cos () sin() cos() ( sin() cos())( cos () ) 6 cos () Recall that we substituted x + = cos(), which we can rewrite as cos() = x+ adj hyp =. If we imagine a triangle in which the adjacent side is x + and the hypotenuse is, then the opposide side must be 4 (x + ), or x x. x x x + Therefore: sin() = x x sin() cos() ( sin() cos())( cos () ) 6 cos () x x x + x x x + ( )(( x + x+ ) 6( ) ) (x + ) x x (x + ) x x (x + ) ( ) 6 (x + ) = (x + ) x x (x + ) You can find all my notes at http://omgimanerd.tech/notes. If you have any questions, comments, or concerns, please contact me at alvin@omgimanerd.tech 5