∫sin x / x = ∫(1 - (x^{2}/3!) + (x^{4}/5!) - (x^{6}/7!) + ....)dx

We know that, ∫x^{n}dx = x^{(n+1)}/n+1 + C

where, C is the integration constant

Thus,

∫sin x / x = x - (x^{3}/3×3!) + (x^{5}/5×5!) - (x^{7}/7×7! ) + ....+ C

We can make use of taylor series of expansion.

Let's look into the stepwise solution

**Explanation:**

Let us see the taylor series of expansion for sin x.

We have,

sin x = x - (x^{3}/3!) + (x^{5}/5!) - (x^{7}/7!) + ..... upto infinite terms. [where ! sign denotes the factorial of a number]

sin x / x = 1/x { x - (x^{3}/3!) + (x^{5}/5!) - (x^{7}/7!) + .....}

= 1 - (x^{2}/3!) + (x^{4}/5!) - (x^{6}/7!) + ....

Now,

∫sin x / x = ∫(1 - (x^{2}/3!) + (x^{4}/5!) - (x^{6}/7!) + ....)dx

We know that, ∫x^{n}dx = x^{(n+1)}/n+1 + C

where, C is the integration constant

Thus,

∫sin x / x = x - (x^{3}/3×3!) + (x^{5}/5×5!) - (x^{7}/7×7! ) + ....+ C