These trig identities may be useful to know.

Pythagorean Identities (PI):


sin2⁡(x)+cos2⁡(x)=1{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1}


sin2⁡(x)=1−cos2⁡(x){\displaystyle \sin ^{2}(x)=1-\cos ^{2}(x)}


cos2⁡(x)=1−sin2⁡(x){\displaystyle \cos ^{2}(x)=1-\sin ^{2}(x)}


sec2⁡(x)=tan2⁡(x)+1{\displaystyle \sec ^{2}(x)=\tan ^{2}(x)+1}


csc2⁡(x)=cot2⁡(x)+1{\displaystyle \csc ^{2}(x)=\cot ^{2}(x)+1}



Double Angle Identities (DAI):


sin⁡(2x)=2sin⁡(x)cos⁡(x){\displaystyle \sin(2x)=2\sin(x)\cos(x)}


cos⁡(2x)=cos2⁡(x)−sin2⁡(x){\displaystyle \cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)}


cos⁡(2x)=1−sin2⁡(x){\displaystyle \cos(2x)=1-\sin ^{2}(x)}


cos2⁡(x)=12(1+cos⁡(2x)){\displaystyle \cos ^{2}(x)={\frac {1}{2}}(1+\cos(2x))}


sin2⁡(x)=12(1−cos⁡(2x)){\displaystyle \sin ^{2}(x)={\frac {1}{2}}(1-\cos(2x))}


;
,,,,, This is done to simplify the integrand.

Let u=ln⁡(x){\displaystyle u=\ln(x)}. Then du=1xdx{\displaystyle du={\frac {1}{x}}dx}. This is done to simplify the integrand. 

Substitution yields ∫sin5⁡(u)du{\displaystyle \int \sin ^{5}(u)du}


, The goal here is to get a term defined by a trig identity. 


∫sin4⁡(u)sin⁡(u)du{\displaystyle \int \sin ^{4}(u)\sin(u)du}


∫(sin2⁡(u))2sin⁡(u)du{\displaystyle \int {(\sin ^{2}(u))}^{2}\sin(u)du}


, Our goal here is to set up the integrand for another u-sub. Therefore, we want the integrand to consist of a trig function and it's known derivative. 


∫(1−cos2⁡(u))2sin⁡(u)du{\displaystyle \int {(1-\cos ^{2}(u))}^{2}\sin(u)du}


, 

Let w=cos⁡(u){\displaystyle w=\cos(u)}. Then −dw=sin⁡(u)du{\displaystyle -dw=\sin(u)du}

Substitution yields −∫(1−w2)2dw{\displaystyle -\int {(1-w^{2})}^{2}dw}


, Distribute the two binomials. Combine like terms.


−∫1−2w2+w4dw{\displaystyle -\int 1-2w^{2}+w^{4}dw}


,


−+C{\displaystyle -+C}


, 


w=cos⁡(u){\displaystyle w=\cos(u)} and u=ln⁡(x){\displaystyle u=\ln(x)}


−+C{\displaystyle -+C}


,,,, This is done to simplify the integrand. 

 Let u=7x{\displaystyle u=7x}. Then 17du=dx{\displaystyle {\frac {1}{7}}du=dx}.

Substitution yields 17∫sin2⁡(u)cos2⁡(u)du{\displaystyle {\frac {1}{7}}\int \sin ^{2}(u)\cos ^{2}(u)du}


, The goal here is to get the integrand in terms of one trig function.


∫du{\displaystyle \int {}du}


, Simplify.


128∫(1−cos⁡(2u))(1+cos⁡(2u))du{\displaystyle {\frac {1}{28}}\int {(1-\cos(2u))(1+\cos(2u))}du}


128∫1−cos2⁡(2u)du{\displaystyle {\frac {1}{28}}\int {1-\cos ^{2}(2u)}du}


, This is done to make the integrand integrable. 


128∫1−(12(1+cos⁡(4u)))du{\displaystyle {\frac {1}{28}}\int {1-({\frac {1}{2}}(1+\cos(4u)))}du}


, Distribute and combine like terms.


128∫12−12cos⁡(4u)du{\displaystyle {\frac {1}{28}}\int {{\frac {1}{2}}-{\frac {1}{2}}\cos(4u)}du}


,


u=7x{\displaystyle u=7x}


128+C{\displaystyle {\frac {1}{28}}+C}


,,,,, This is done to simplify the integrand. 

Let u=x2{\displaystyle u={\frac {x}{2}}}. Then 2du=dx{\displaystyle 2du=dx}

Change the bounds of integration since the integral at hand is a definite integral. 

When x=0{\displaystyle x=0}, u=0{\displaystyle u=0}

When x=π{\displaystyle x=\pi }, u=π2{\displaystyle u={\frac {\pi }{2}}}

Substitution yields 2∫0π2cos4⁡(u)du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}\cos ^{4}(u)du}


, Get the integrand in to a form such that an identity can be used.


2∫0π2(cos2⁡(u))2du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}{(\cos ^{2}(u))}^{2}du}


,, Simplify. 


2∫0π212(1+cos⁡(2u)+cos2⁡(2u))du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}{\frac {1}{2}}(1+\cos(2u)+\cos ^{2}(2u))du}


, This is done to get cos2⁡(2u){\displaystyle \cos ^{2}(2u)} into a term we can integrate. 


12∫0π21+2cos⁡(2u)+12(1+2cos⁡(4u))du{\displaystyle {\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}1+2\cos(2u)+{\frac {1}{2}}(1+2\cos(4u))du}


, Simplify. 


12∫0π232+2cos⁡(2u)+12cos⁡(4u)du{\displaystyle {\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}{\frac {3}{2}}+2\cos(2u)+{\frac {1}{2}}\cos(4u)du}


, Since the bounds were converted, don't re-sub.  


120π2{\displaystyle {\frac {1}{2}}{}_{0}^{\frac {\pi }{2}}}


,


=3π8{\displaystyle ={\frac {3\pi }{8}}}


,,,,, This is done to simplify the integrand.

 Let u=y{\displaystyle u={\sqrt {y}}}. Then 2du=1ydy{\displaystyle 2du={\frac {1}{\sqrt {y}}}dy}. 

Substitution yields 2∫tan4⁡(u)du{\displaystyle 2\int \tan ^{4}(u)du}


, The goal here is to get a term we can replace with a Pythagorean Identity. 

 2∫tan2⁡(u)tan2⁡(u)du{\displaystyle 2\int \tan ^{2}(u)\tan ^{2}(u)du}


, Only substitute one tan2⁡(u){\displaystyle \tan ^{2}(u)}. This is done to get a term and it's derivative in the integrand; this a set up for a u-sub. 

 2∫(sec2⁡(u)−1)tan2⁡(u)du{\displaystyle 2\int (\sec ^{2}(u)-1)\tan ^{2}(u)du}


, Separate the integrand to make anti-differentiation easier. 

 2∫tan2⁡(u)sec2⁡(u)du−2∫tan2⁡(u)du{\displaystyle 2\int {\tan ^{2}(u)\sec ^{2}(u)}du-2\int {\tan ^{2}(u)}du}


, Then dw=sec2⁡(u)du{\displaystyle dw=\sec ^{2}(u)du}

Substitution yields 2∫w2dw+2∫tan2⁡(u)du{\displaystyle 2\int w^{2}dw+2\int \tan ^{2}(u)du} 


, 

 ∫w2dw+∫sec2⁡(u)−1du{\displaystyle \int w^{2}dw+\int \sec ^{2}(u)-1du}


,


u=y{\displaystyle u={\sqrt {y}}}


23tan3⁡(y)−2tan⁡(y)−2y+C{\displaystyle {\frac {2}{3}}\tan ^{3}({\sqrt {y}})-2\tan({\sqrt {y}})-2{\sqrt {y}}+C}


,,,,, 

Let u=1θ{\displaystyle u={\frac {1}{\theta }}}. Then −du=1θ2{\displaystyle -du={\frac {1}{\theta ^{2}}}}.

Substitution yields −∫sec4⁡(u)tan3⁡(u)du{\displaystyle -\int \sec ^{4}(u)\tan ^{3}(u)du}


, Rearrange such that we can replace a term with a Pythagorean Identity. Note, sec⁡(u)tan⁡(u){\displaystyle \sec(u)\tan(u)} is the elementary derivative of sec⁡(u){\displaystyle \sec(u)}.


−∫sec3⁡(u)tan2⁡(u)sec⁡(u)tan⁡(u)du{\displaystyle -\int \sec ^{3}(u)\tan ^{2}(u)\sec(u)\tan(u)du}


,, Keep sec⁡(u)tan⁡(u){\displaystyle \sec(u)\tan(u)} intact. Remember, we want to have a trig function and it's derivative in the integrand to employ a u-sub. 


∫(sec5⁡(u)−sec3⁡(u))sec⁡(u)tan⁡(u)du{\displaystyle \int (\sec ^{5}(u)-\sec ^{3}(u))\sec(u)\tan(u)du}


,

Let w=sec⁡(u){\displaystyle w=\sec(u)}. Then dw=sec⁡(u)tan⁡(u)du{\displaystyle dw=\sec(u)\tan(u)du}

Substitution yields −∫(w5−w3)dw{\displaystyle -\int (w^{5}-w^{3})dw}


,


−+C{\displaystyle -+C}


,


−+C{\displaystyle -+C}


,,,,, 

Let u=t2{\displaystyle u=t^{2}}. Then du2=tdt{\displaystyle {\frac {du}{2}}=tdt}

Substitution yields 12∫sec4⁡(u)du{\displaystyle {\frac {1}{2}}\int \sec ^{4}(u)du}


, Rearrange such that we can use a Pythagorean Identity. 


12∫sec2⁡(u)sec2⁡(u)du{\displaystyle {\frac {1}{2}}\int \sec ^{2}(u)\sec ^{2}(u)du}


, Keep one sec2⁡(u){\displaystyle \sec ^{2}(u)} intact. We will need for a u-sub. 


12∫(1+tan2⁡(u))sec2⁡(u)du{\displaystyle {\frac {1}{2}}\int (1+\tan ^{2}(u))\sec ^{2}(u)du}


,

 Let w=tan⁡(u){\displaystyle w=\tan(u)}. Then dw=sec2⁡(u){\displaystyle dw=\sec ^{2}(u)}

Substitution yields 12∫1+w2dw{\displaystyle {\frac {1}{2}}\int 1+w^{2}dw}


,


12+C{\displaystyle {\frac {1}{2}}+C}


, 


w=tan⁡(u){\displaystyle w=\tan(u)} and u=t2{\displaystyle u=t^{2}}


tan⁡(t2)2+tan⁡(t2)6+C{\displaystyle {\frac {\tan(t^{2})}{2}}+{\frac {\tan(t^{2})}{6}}+C}
Know the relevant trig identities.
Separate a sin2⁡(x){\displaystyle \sin ^{2}(x)}.
,
Rewrite all the rest in terms of cos⁡(x){\displaystyle \cos(x)} and/or sin⁡(x){\displaystyle \sin(x)} using Pythagorean Identities.
,
Use algebra, substitution, and trig identities when applicable.
,
Evaluate ∫sin5⁡(lnx)xdx{\displaystyle \int {\frac {\sin ^{5}(lnx)}{x}}dx}.
,
Do a u-sub.
Manipulate the integrand algebraically.
Use a Pythagorean Identity to get sin2⁡(u){\displaystyle \sin ^{2}(u)} in terms of cosine.
Do a u-sub.
Manipulate the integrand algebraically.
Integrate.
Re-sub.Use a Double-Angle Identity for sin2⁡(x){\displaystyle \sin ^{2}(x)} and/or cos2⁡(x){\displaystyle \cos ^{2}(x)}
,
Use algebra, substitution, and trig identities when applicable. 
,
Evaluate ∫sin2⁡(7x)cos2⁡(7x)dx{\displaystyle \int \sin ^{2}(7x)\cos ^{2}(7x)dx}.
,
Do a u-sub.
Use a  Double-Angle Identity for sin2⁡(u){\displaystyle \sin ^{2}(u)} and cos2⁡(u){\displaystyle \cos ^{2}(u)}.
Manipulate the integrand algebraically.
Use a Double-Angle Identity for cos2⁡(2u){\displaystyle \cos ^{2}(2u)}.
Manipulate the integrand algebraically.
Integrate and re-sub.
Separate a sin2⁡(x){\displaystyle \sin ^{2}(x)}.
,
Rewrite all the rest in terms of cos⁡(x){\displaystyle \cos(x)} and/or sin⁡(x){\displaystyle \sin(x)} using Pythagorean Identities.
,
Use algebra, substitution, and trig identities when applicable.
,Evaluate ∫πcos4x2dx{\displaystyle \int _{0}^{\pi }\cos ^{4}{\frac {x}{2}}dx}.
,
Do a u-sub.
Manipulate the integrand algebraically.
Use a Double-Angle Identity for cos2⁡(u){\displaystyle \cos ^{2}(u)}


2∫0π22du{\displaystyle 2\int _{0}^{\frac {\pi }{2}}{}^{2}du}


,
Manipulate the integrand algebraically.
Use a Double-Angle Identity for cos2⁡(2u){\displaystyle \cos ^{2}(2u)}.
Manipulate the integrand algebraically.
Integrate.
Compute the integral.
Separate a tan⁡(x){\displaystyle \tan(x)}.
,
Use Pythagorean Identities to get all else in terms of secant.
,
Use algebra, substitution, and trig identities when applicable. 
,
Evaluate ∫tan4yydy{\displaystyle \int {\frac {\tan ^{4}{\sqrt {y}}}{\sqrt {y}}}dy}.
,
Do a u-sub.
Manipulate the integrand algebraically.
Use a Pythagorean Identity for tan2⁡(x){\displaystyle \tan ^{2}(x)}.
Manipulate the integrand algebraically.
Do a u-sub on the first integral

 Let w=tan⁡(u){\displaystyle w=\tan(u)}.
Use a Pythagorean Identity on the second integral.
Integrate and re-sub.
Separate a tan⁡(x){\displaystyle \tan(x)}.
,
Use Pythagorean Identities to get all else in terms of secant.
,
Substitute a sec⁡(x){\displaystyle \sec(x)}. 
,Evaluate ∫sec4⁡(1θ)tan3⁡(1θ)1θ2dθ{\displaystyle \int \sec ^{4}({\frac {1}{\theta }})\tan ^{3}({\frac {1}{\theta }}){\frac {1}{\theta ^{2}}}d\theta }
,
Do a u-sub.
Manipulate the integrand algebraically.
Use a Pythagorean Identity for tan2⁡(u){\displaystyle \tan ^{2}(u)}


∫sec3⁡(u)(sec2⁡(u)−1)sec⁡(u)tan⁡(u)du{\displaystyle \int \sec ^{3}(u)(\sec ^{2}(u)-1)\sec(u)\tan(u)du}


,
Manipulate the integrand algebraically.
Do a u-sub.
Integrate.
Re-sub.
Separate a sec2⁡(x){\displaystyle \sec ^{2}(x)}.
,
Use a Pythagorean Identity to get all else in terms of tangent.
,
Substitute a tan⁡(x){\displaystyle \tan(x)} when applicable.
,Evaluate ∫tsec4⁡(t2)dt{\displaystyle \int t\sec ^{4}(t^{2})dt}
,
Do a u-sub.
Manipulate the integrand algebraically.
Use a Pythagorean Identity for sec2⁡(u){\displaystyle \sec ^{2}(u)}.
Do a u-sub.
Integrate.
Re-sub.
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