{"id":10269,"date":"2017-06-19T15:10:20","date_gmt":"2017-06-19T22:10:20","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=10269"},"modified":"2017-06-18T16:13:38","modified_gmt":"2017-06-18T23:13:38","slug":"ap-calculus-bc-review-integration-substitution","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-bc-review-integration-substitution\/","title":{"rendered":"AP Calculus BC Review: Integration By Substitution"},"content":{"rendered":"<p><strong>Substitution<\/strong> is just one of the many techniques available for finding <em>indefinite integrals<\/em> (that is, <em>antiderivatives<\/em>). Let&#8217;s review the method of integration by substitution and get some practice for the AP Calculus BC exam.<\/p>\n<h2>The Substitution Rule<\/h2>\n<p>Integration by substitution, also known as <strong><em>u<\/em>-substitution<\/strong>, after the most common variable for substituting, allows you to reduce a complicated integral to one that is easier to work with.<\/p>\n<p>The formula works as follows. Suppose that <em>F<\/em> is an antiderivative for <em>f<\/em>. Then we have:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-8746\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/01\/Substitution_rule.gif\" alt=\"Integration by Substitution Rule\" width=\"547\" height=\"41\" \/><\/p>\n<p>In fact, you can think of the Substitution Rule as reversing the <em>Chain Rule<\/em>.<\/p>\n<p>Basically, this rule states that if you have a complicated integral like the one on the left, then it instantly reduces to a simpler one that can be worked out with no trouble.<\/p>\n<div align=\"center\">\n<figure id=\"attachment_9567\" aria-describedby=\"caption-attachment-9567\" style=\"width: 468px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\" wp-image-9567\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/cat-633081_640-600x450.jpg\" alt=\"surprised cat\" width=\"468\" height=\"351\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/cat-633081_640-600x450.jpg 600w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/cat-633081_640-300x225.jpg 300w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/cat-633081_640-30x23.jpg 30w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/cat-633081_640.jpg 640w\" sizes=\"(max-width: 468px) 100vw, 468px\" \/><figcaption id=\"caption-attachment-9567\" class=\"wp-caption-text\">Okay, I&#8217;m listening.<\/figcaption><\/figure>\n<\/div>\n<p>However, the hard part is arranging a given integral in the right way. That&#8217;s where the step-by-step method outlined below comes in.<\/p>\n<p>Furthermore, substituting isn&#8217;t a magic bullet that can tackle every integral. Just think of it as one powerful tool in your toolbox. For more tools and information about integration, check out the following resource. <a href=\"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-review-indefinite-integrals\/\">AP Calculus Review: Indefinite Integrals<\/a>.<\/p>\n<h2>The Method of Integration by Substitution<\/h2>\n<p>Next let&#8217;s review the main steps in <em>u<\/em>-substitution.<\/p>\n<h3>Step 1. Choose Your Substitution<\/h3>\n<p>If you want to use substitution, then the first thing to do is to identify what you want to substitute.<\/p>\n<p>In other words, you have to make a choice for what <em>u<\/em> = <em>g<\/em>(<em>x<\/em>) will be in your integral.<\/p>\n<p>But what should you choose? That comes with experience.<\/p>\n<p>There are a few good rules of thumb to follow when choosing <em>u<\/em>, but these are by no means fool-proof. So if the first choice doesn&#8217;t work, try something else.<\/p>\n<ul>\n<li>Expression within a set of parentheses<\/li>\n<li>Expression within a radical<\/li>\n<li>Denominator (if the integrand is a single fractional expression)<\/li>\n<li>Either sin <em>x<\/em> or cos <em>x<\/em> if the other trig function is also present<\/li>\n<li>Exponent of <em>e<\/em><\/li>\n<li>Trig (reverse) Substitution<\/li>\n<\/ul>\n<h3>Step 2. Find the Differential<\/h3>\n<p>Once you&#8217;ve decided on your substitution, the next step is to find its<strong> differential<\/strong>. Remember, the differential of a function of <em>x<\/em> is just its derivative times <em>dx<\/em>.<\/p>\n<p>So the differential of <em>u<\/em> = <em>g<\/em>(<em>x<\/em>) is <em>du<\/em> = <em>g<\/em>\u00a0&#8216;(<em>x<\/em>)\u00a0<em>dx<\/em>,<\/p>\n<p>The reason you should do this step is that the differential provides a link between the &#8220;<em>dx<\/em>&#8221; from the original integral and the new integral, which will have &#8220;<em>du<\/em>&#8221; instead.<\/p>\n<h3>Step 3. Rewrite the Integral<\/h3>\n<p>Now you get to change the form of the integral.<\/p>\n<p>Replace <em>g<\/em>(<em>x<\/em>) by <em>u<\/em>, hopefully making the integrand a little simpler.<\/p>\n<p>But don&#8217;t be worried if there are still expressions in the integral involving <em>x<\/em>. In fact, those leftover expressions can be traded out using <em>du<\/em> = <em>g<\/em>\u00a0&#8216;(<em>x<\/em>)\u00a0<em>dx<\/em>.<\/p>\n<p>The only catch is that you have to have a <em>perfect<\/em> match in order to make the trade. We&#8217;ll see more about this point in the examples below.<\/p>\n<h3>Step 4. Simplify and Integrate<\/h3>\n<p>Next, check to see what variables are represented in the integral. If there are any leftover <em>x<\/em>&#8216;s anywhere, or if the old differential <em>dx<\/em> is still in the integral, then the substitution is not complete.<\/p>\n<p>However if there are only expressions of <em>u<\/em>, and if the differential is <em>du<\/em>, then you can move on to the integration step.<\/p>\n<p>Simplify the integrand as much as possible. Hopefully the result is a much simpler integral (in the variable <em>u<\/em>) than the one you started with (in <em>x<\/em>).<\/p>\n<p>Now is the time to integrate!<\/p>\n<figure id=\"attachment_10293\" aria-describedby=\"caption-attachment-10293\" style=\"width: 473px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\" wp-image-10293\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/4378302122_bf5782d121_z-600x398.jpg\" alt=\"Handy Manny and Bob the Builder\" width=\"473\" height=\"314\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/4378302122_bf5782d121_z-600x398.jpg 600w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/4378302122_bf5782d121_z-300x199.jpg 300w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/4378302122_bf5782d121_z.jpg 640w\" sizes=\"(max-width: 473px) 100vw, 473px\" \/><figcaption id=\"caption-attachment-10293\" class=\"wp-caption-text\">Can we integrate it? Yes we can! <em>Image by <a href=\"https:\/\/www.flickr.com\/photos\/jdhancock\/\" target=\"_blank\" rel=\"noopener noreferrer\">JD Hancock<\/a><\/em>.<\/figcaption><\/figure>\n<h3>Step 5. Replace the Substitution<\/h3>\n<p>Assuming you made it this far, you now have a function of the variable <em>u<\/em> (plus a constant of integration). The final step is to replace every <em>u<\/em> by its expression in <em>x<\/em>, namely the function <em>g<\/em>(<em>x<\/em>) that you chose back in Step 1.<\/p>\n<h2>Examples<\/h2>\n<p>Ok let&#8217;s see if we can use the method of integration by substitution in a few examples.<\/p>\n<h3>Example 1 \u2014 An Exponential Experience<\/h3>\n<p><img decoding=\"async\" class=\"alignnone size-full wp-image-10298\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex1.gif\" alt=\"integral of x e^(x^2)\" width=\"96\" height=\"41\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex1.gif 96w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex1-30x13.gif 30w\" sizes=\"(max-width: 96px) 100vw, 96px\" \/><\/p>\n<h4>Solution<\/h4>\n<p>The best choice for substitution would be the exponent of <em>e<\/em>. That is, <em>u<\/em> = <em>x<\/em><sup>2<\/sup>.<\/p>\n<p>Then the differential would be: <em>du<\/em> = 2<em>x<\/em>\u00a0<em>dx<\/em>.<\/p>\n<p>However there is no constant 2 in the original integral. The best way I&#8217;ve found to deal with mismatched constants is to divide them from both sides of the differential: <img decoding=\"async\" class=\"alignnone size-full wp-image-10299\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/differential1.gif\" alt=\"1\/2 du = x dx\" width=\"89\" height=\"37\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/differential1.gif 89w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/differential1-30x12.gif 30w\" sizes=\"(max-width: 89px) 100vw, 89px\" \/>.<\/p>\n<p>Then we can make our substitutions and solve the integral. Don&#8217;t forget to replace your substitution at the end!<\/p>\n<p>I&#8217;ve used blue for the <em>u<\/em>-substitution and red for its differential to make the process a little clearer.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10302\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex1_solution.gif\" alt=\"Solution: integral is 1\/2 e^(x^2) + C\" width=\"182\" height=\"126\" \/><\/p>\n<h3>Example 2 \u2014 A Bit <em>Trig<\/em>-ier<\/h3>\n<p><img decoding=\"async\" class=\"alignnone size-full wp-image-10303\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex2.gif\" alt=\"integral of sin^6 x cos^3 x\" width=\"153\" height=\"41\" \/><\/p>\n<h4>Solution<\/h4>\n<p>This one is tricky. In fact, a simple integration by substitution won&#8217;t work. Not until we rewrite the integrand using a trig identity, that is.<\/p>\n<p>Remember the <em>Pythagorean Identity<\/em>? Well it can be used to exchange squared powers of sine for cosine and vice versa.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-9520\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_identity.gif\" alt=\"Rewriting the Pythagorean Identity\" width=\"333\" height=\"17\" \/><\/p>\n<p>The key step here is to isolate a single factor of cos <em>x<\/em> and convert the remaining factor into an expression of sin <em>x<\/em> using the Pythagorean Identity.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10304\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/ex2_trig_rewrite.gif\" alt=\"rewriting the integrand using the Pythagorean Identity\" width=\"385\" height=\"134\" \/><\/p>\n<p>Now we can see what the substitution will be. Let <em>u<\/em> = sin <em>x<\/em>.<\/p>\n<p>Then the corresponding differential is: <em>du<\/em> = cos <em>x<\/em> <em>dx<\/em>.<\/p>\n<p>Fortunately, there is a perfect match for the differential this time. So the rest of the process is fairly straightforward. Just remember that sin<em><sup>n<\/sup><\/em> <em>x<\/em> stands for (sin <em>x<\/em>)<em><sup>n<\/sup><\/em>.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10305\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex2_solution.gif\" alt=\"Solution to example. (sin^7 x) \/ 7 - (sin^9 x) \/ 9 + C\" width=\"395\" height=\"132\" \/><\/p>\n<h3>Example 3 \u2014 Trig Substitution (Challenging)<\/h3>\n<p><img decoding=\"async\" class=\"alignnone size-full wp-image-10306\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex3.gif\" alt=\"integral of 1\/(4+x^2)^(3\/2)\" width=\"125\" height=\"42\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex3.gif 125w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex3-30x10.gif 30w\" sizes=\"(max-width: 125px) 100vw, 125px\" \/><\/p>\n<h4>What&#8217;s Our Gameplan?<\/h4>\n<p>As far as substitution problems go, this one is by far the hardest type that you may find on the AP Calculus BC test.<\/p>\n<p>The first natural choice for substitution, <em>u<\/em> = 4 + <em>x<\/em><sup>2<\/sup>, won&#8217;t work. The reason is that the differential, <em>du<\/em> = 2<em>x<\/em> <em>dx<\/em>, has that extra &#8220;x&#8221; in it that cannot be matched in the integrand.<\/p>\n<p>Instead, we will make something like a <em>reverse substitution<\/em>, called a <strong>trig substitution<\/strong>.<\/p>\n<p>When you have an expression like <em>a<\/em><sup>2<\/sup> + <em>x<\/em><sup>2<\/sup>, then the trig substitution <em>x<\/em> = <em>a<\/em> tan <em>\u03b8<\/em> often helps.<\/p>\n<p>Why? Because of the identity, 1 + tan<sup>2<\/sup> <em>\u03b8<\/em> = sec<sup>2<\/sup> <em>\u03b8<\/em>.<\/p>\n<p>(Note, there are two other kinds of trig substitution. For more details, check out this <a href=\"https:\/\/www.khanacademy.org\/math\/calculus-home\/integration-techniques-calc\/trigonometric-substitution-calc\/v\/introduction-to-trigonometric-substitution\" target=\"_blank\" rel=\"noopener noreferrer\">video<\/a>.)<\/p>\n<p>Let&#8217;s see how a trig substitution can help us out in this situation.<\/p>\n<h4>Solution<\/h4>\n<p>Substitute <em>x<\/em> = 2 tan <em>\u03b8<\/em><\/p>\n<p>The differential is: <em>dx<\/em> = 2 sec<sup>2<\/sup> <em>\u03b8<\/em>\u00a0<em>d\u03b8<\/em>, which tells us exactly how to replace <em>dx<\/em> in the original integral. <em>Warning:<\/em> this could get messy!<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10307\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex3_solution_partA.gif\" alt=\"Partial solution to the trig substitution problem\" width=\"308\" height=\"291\" \/><\/p>\n<p>Now after all that work, we&#8217;re in the home stretch! Simply rewrite the reciprocal of secant in a more useful form and integrate.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10308\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex3_solution_partB.gif\" alt=\"integrating 1 over sec theta\" width=\"326\" height=\"41\" \/><\/p>\n<p>Finally, we have to reinterpret the answer in the original variable <em>x<\/em>.<\/p>\n<p>The best way to do this is to think <em>trigonometrically<\/em>. Based on our chosen substitution, <em>x<\/em> = 2 tan <em>\u03b8<\/em>, we know that tan <em>\u03b8<\/em> = <em>x<\/em>\/2.<\/p>\n<p>Interpret this information in a right triangle. The opposite side could be <em>x<\/em> while the adjacent side is 2. That makes the hypotenuse equal to: <img decoding=\"async\" class=\"alignnone size-full wp-image-10309\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/sqrt.gif\" alt=\"square root of 4 + x^2\" width=\"63\" height=\"19\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/sqrt.gif 63w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/sqrt-30x9.gif 30w\" sizes=\"(max-width: 63px) 100vw, 63px\" \/> (Doesn&#8217;t that expression look familiar?)<\/p>\n<p>Therefore, sin <em>\u03b8<\/em>, which is opposite over hypotenuse, is equal to:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10311\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/sin_theta.gif\" alt=\"sin theta\" width=\"124\" height=\"38\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/sin_theta.gif 124w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/sin_theta-30x9.gif 30w\" sizes=\"(max-width: 124px) 100vw, 124px\" \/><\/p>\n<p>Putting it all together,<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-10313\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/substitution_ex3_solution.gif\" alt=\"Final Solution for Example 3\" width=\"353\" height=\"43\" \/><\/p>\n<h2>Conclusion<\/h2>\n<p>Integration by substitution is a powerful tool that gets used quite often in practice. It&#8217;s worth your time to master this technique in order to score as high as possible on the AP Calculus BC exam.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Substitution is just one of the many techniques available for finding indefinite integrals (that is, antiderivatives). Let&#8217;s review the method of integration by substitution and get some practice for the AP Calculus BC exam. The Substitution Rule Integration by substitution, also known as u-substitution, after the most common variable for substituting, allows you to reduce [&hellip;]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-10269","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>AP Calculus BC Review: Integration By Substitution - Magoosh Blog | High School<\/title>\n<meta name=\"description\" content=\"Integration by substitution is just one of the many techniques available for finding integrals (or antiderivatives). 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