IntegralTak Tentu Mari lanjutkan membahas integral sedikit lebih jauh.Perhatikan table berikut ini Fungsi Turunannya 𝑦 = 𝑥2 + 2 𝑦 ′ = 2𝑥 𝑦 = 𝑥2 + 5 𝑦 ′ = 2𝑥 𝑦 = 𝑥 2 + 10 𝑦 ′ = 2𝑥 𝑦 = 𝑥 2 − 20 𝑦 ′ = 2𝑥 Fungsi yang berbeda-beda pada kolom 1, menghasilkan turunan yang sama pada kolom 2. Contoh4. ∫ (x2+ 1)5.2x dx = (x2+ 1)6/6 + C. (Disini kita menerapkan Aturan Pangkat yang Diperumum dengan g(x) = x2 + 1, g'(x) = 2x.) Contoh 5. Jika g(x) = sin x, maka g'(x) = cos x. Jadi, menurut Aturan Pangkat yang Diperumum, diperoleh ∫ sin dx = (sin x)2/2 + C. Latihan. Tentukan integral tak tentu di bawah ini. 1. ∫(x2+ x-2 ContohSoal 5. Diperoleh a db ab b da a db 2x sin x sin x. 1 2 x 3 sin 2x π 1 2 sin 2x π dx 1 2 x 3 sin 2x π 1 2 1 2 cos 2x π 1 2 x 3 sin 2x π 1 4 cos 2x π kalikan 16 tambahkan C nya 16 1 2 x 3 sin 2x π 1 4 cos 2x π C 8 x 3 sin 2x π 4 cos 2x π C. Berikut ini kami berikan contoh-contoh soal integral yang menggunakan logaritma natural. ContohSoal Integral Substitusi. Berikut ini adalah contoh soal integral substitusi aljabar beserta dengan pembahasannya, simak baik-baik ya! Tentukanlah integral dari. Nah untuk menjawab soal integral di atas, kita ambil pemisalan. Biasanya yang di dalam tanda kurung atau di dalam tanda akar atau yang pangkatnya paling besar. . The answer is =-1/5cos^5x+2/3cos^3x-cosx+C Explanation We need sin^2x+cos^2x=1 The integral is intsin^5dx=int1-cos^2x^2sinxdx Perform the substitution u=cosx, =>, du=-sinxdx Therefore, intsin^5dx=-int1-u^2^2du =-int1-2u^2+u^4du =-intu^4du+2intu^2du-intdu =-u^5/5+2u^3/3-u =-1/5cos^5x+2/3cos^3x-cosx+C This integral is mostly about clever rewriting of your functions. As a rule of thumb, if the power is even, we use the double angle formula. The double angle formula says sin^2theta=1/21-cos2theta If we split up our integral like this, int\ sin^2x*sin^2x\ dx We can use the double angle formula twice int\ 1/21-cos2x*1/21-cos2x\ dx Both parts are the same, so we can just put it as a square int\ 1/21-cos2x^2\ dx Expanding, we get int\ 1/41-2cos2x+cos^22x\ dx We can then use the other double angle formula cos^2theta=1/21+cos2theta to rewrite the last term as follows 1/4int\ 1-2cos2x+1/21+cos4x\ dx= =1/4int\ 1\ dx-int\ 2cos2x\ dx+1/2int\ 1+cos4x\ dx= =1/4x-int\ 2cos2x\ dx+1/2x+int\ cos4x\ dx I will call the left integral in the parenthesis Integral 1, and the right on Integral 2. Integral 1 int\ 2cos2x\ dx Looking at the integral, we have the derivative of the inside, 2 outside of the function, and this should immediately ring a bell that you should use u-substitution. If we let u=2x, the derivative becomes 2, so we divide through by 2 to integrate with respect to u int\ cancel2cosu/cancel2\ du int\ cosu\ du=sinu=sin2x Integral 2 int\ cos4x\ dx It's not as obvious here, but we can also use u-substitution here. We can let u=4x, and the derivative will be 4 1/4int\ cosu\ dx=1/4sinu=1/4sin4x Completing the original integral Now that we know Integral 1 and Integral 2, we can plug them back into our original expression to get the final answer 1/4x-sin2x+1/2x+1/4sin4x+C= =1/4x-sin2x+1/2x+1/8sin4x+C= =1/4x-1/4sin2x+1/8x+1/32sin4x+C= =3/8x-1/4sin2x+1/32sin4x+C

integral sin pangkat 5 x dx