Standard Integration Techniques Pas encore traduit La version anglaise s’affiche pour l’instant. Cette page passera en français dès que la traduction sera prête. U Substitution
The u substitution method is useful for integrals involving a composite function. It follows the principle that:
∫ f ( g ( x ) ) ⋅ g ′ ( x ) d x = ∫ f ( u ) d u \int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du ∫ f ( g ( x )) ⋅ g ′ ( x ) d x = ∫ f ( u ) d u
where u = g ( x ) u = g(x) u = g ( x ) and d u = g ′ ( x ) d x du = g'(x)dx d u = g ′ ( x ) d x . This technique is also applicable without limits for indefinite integrals.
Example
For the integral ∫ 1 2 5 x 2 cos ( x 3 ) d x \int_1^2 5x^2 \cos(x^3) \, dx ∫ 1 2 5 x 2 cos ( x 3 ) d x , we set u = x 3 u = x^3 u = x 3 which gives us d u = 3 x 2 d x du = 3x^2 \, dx d u = 3 x 2 d x . The limits are also changed accordingly to u u u values. The integral simplifies to:
∫ 1 8 5 3 cos ( u ) d u = [ 5 3 sin ( u ) ] 1 8 = 5 3 ( sin ( 8 ) − sin ( 1 ) ) \int_1^8 \frac{5}{3} \cos(u) \, du = \left[ \frac{5}{3} \sin(u) \right]_1^8 = \frac{5}{3} (\sin(8) - \sin(1)) ∫ 1 8 3 5 cos ( u ) d u = [ 3 5 sin ( u ) ] 1 8 = 3 5 ( sin ( 8 ) − sin ( 1 ))
Products and Quotients of Trig Functions
For Products
When integrating products of sine and cosine, consider the following strategies based on the powers of sine and cosine:
If the power of sine is odd, move one sine out and convert the rest to cosines using sin 2 ( x ) = 1 − cos 2 ( x ) \sin^2(x) = 1 - \cos^2(x) sin 2 ( x ) = 1 − cos 2 ( x ) .
If the power of cosine is odd, move one cosine out and convert the rest to sines using cos 2 ( x ) = 1 − sin 2 ( x ) \cos^2(x) = 1 - \sin^2(x) cos 2 ( x ) = 1 − sin 2 ( x ) .
If both powers are odd, use the above strategies.
If both powers are even, use double angle or half angle formulas.
For Quotients
When integrating products of tangent and secant, the strategies are as follows:
If the power of tangent is odd, save one tangent and convert the rest to secants.
If the power of secant is even, save one secant and convert the rest to tangents.
Use appropriate substitutions such as u = tan ( x ) u = \tan(x) u = tan ( x ) or u = sec ( x ) u = \sec(x) u = sec ( x ) depending on the scenario.
Example - Products
For the integral ∫ tan 3 ( x ) sec 5 ( x ) d x \int \tan^3(x) \sec^5(x) \, dx ∫ tan 3 ( x ) sec 5 ( x ) d x , use u = sec ( x ) u = \sec(x) u = sec ( x ) , d u = sec ( x ) tan ( x ) d x du = \sec(x)\tan(x) \, dx d u = sec ( x ) tan ( x ) d x . This leads to:
∫ u 4 ( u 2 − 1 ) d u = ∫ ( u 6 − u 4 ) d u \int u^4(u^2 - 1) \, du = \int (u^6 - u^4) \, du ∫ u 4 ( u 2 − 1 ) d u = ∫ ( u 6 − u 4 ) d u
After integration, revert back to x x x :
1 7 sec 7 ( x ) − 1 5 sec 5 ( x ) + C \frac{1}{7} \sec^7(x) - \frac{1}{5} \sec^5(x) + C 7 1 sec 7 ( x ) − 5 1 sec 5 ( x ) + C
Example - Quotients
For the integral ∫ sin 3 ( x ) cos 3 ( x ) d x \int \frac{\sin^3(x)}{\cos^3(x)} \, dx ∫ c o s 3 ( x ) s i n 3 ( x ) d x , use the identity sin 2 ( x ) = 1 − cos 2 ( x ) \sin^2(x) = 1 - \cos^2(x) sin 2 ( x ) = 1 − cos 2 ( x ) and the substitution u = cos ( x ) u = \cos(x) u = cos ( x ) :
∫ ( 1 − cos 2 ( x ) ) sin ( x ) cos 3 ( x ) d x = ∫ ( 1 u 2 − 1 u 4 ) d u \int \frac{(1 - \cos^2(x))\sin(x)}{\cos^3(x)} \, dx = \int \left( \frac{1}{u^2} - \frac{1}{u^4} \right) \, du ∫ c o s 3 ( x ) ( 1 − c o s 2 ( x )) s i n ( x ) d x = ∫ ( u 2 1 − u 4 1 ) d u
After integrating and reverting back to x x x :
1 2 sec 2 ( x ) + 2 ln ∣ cos ( x ) ∣ + 1 2 cos 2 ( x ) + C \frac{1}{2} \sec^2(x) + 2 \ln |\cos(x)| + \frac{1}{2} \cos^2(x) + C 2 1 sec 2 ( x ) + 2 ln ∣ cos ( x ) ∣ + 2 1 cos 2 ( x ) + C
The trigonometric formulas useful in these integrations include:
sin ( 2 x ) = 2 sin ( x ) cos ( x ) \sin(2x) = 2 \sin(x) \cos(x) sin ( 2 x ) = 2 sin ( x ) cos ( x )
cos ( 2 x ) = 1 − 2 sin 2 ( x ) \cos(2x) = 1 - 2 \sin^2(x) cos ( 2 x ) = 1 − 2 sin 2 ( x )
sin 2 ( x ) = 1 2 ( 1 − cos ( 2 x ) ) \sin^2(x) = \frac{1}{2}(1 - \cos(2x)) sin 2 ( x ) = 2 1 ( 1 − cos ( 2 x ))